TERM - II Written as per the syllabus prescribed by the Central Board of Secondary Education. CBSE...
Transcript of TERM - II Written as per the syllabus prescribed by the Central Board of Secondary Education. CBSE...
MATHEMATICS TERM - II
Written as per the syllabus prescribed by the Central Board of Secondary Education.
CBSECLASS X
Printed at: Repro Knowledgecast Ltd., Mumbai
10441_11071_JUP
P.O. No. 34971
© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical
including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.
Salient Features
• Extensive coverage of the syllabus for Term - II in an effortless and easy to grasp format.
• In alignment with the latest paper pattern of Central Board of Secondary Education.
• Neat and labelled diagrams.
• Constructions drawn with accurate measurements.
• ‘Things to Remember’ highlights important facts.
• Variety of additional problems for practice.
• Questions from previous years board papers have been solved.
• Memory Maps at the end of each chapter to facilitate quick revision.
• Sample Test Paper at the end of each chapter designed for student’s Self Assessment.
• Model Question Papers in accordance with the latest paper pattern.
PREFACE In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you. “Class X: Mathematics” is a complete and thorough guide critically analyzed and extensively drafted to foster the student’s confidence. The book ensures extensive coverage of the syllabus for Term - II in an effortless and easy to grasp format.
The Topic-wise classified format for each chapter of this book helps the students in easy comprehension.
Each chapter includes the following features:
Theory of each mathematical concept is explained with appropriate references. NCERT exercises and Exemplar questions are covered with solutions. Problems based on NCERT exercises covering all the topics are provided with solutions. Practice problems based on NCERT exercises and MCQs help students to revise the topics thoroughly. One Mark Questions are provided for each chapter. Higher order thinking skills (HOTS) questions have been added for the student to gain insight on the various levels of theory-based questions. Value - Based Questions which emphasize on values have been included. Memory Map has been provided to give a quick overview of the chapter, helping the students in effective learning. Sample Test Paper at the end of each chapter helps to test the range of preparation of the students. ‘Things to Remember’ help the students’ gain knowledge required to understand different concepts.
Two Model Question Papers, designed as per CBSE Paper Pattern, are a unique tool to enable self-assessment for the students. Answers for previous years Board Questions have been included in this book. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.
Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.
Best of luck to all the aspirants! Yours faithfully,
Publisher
UNIT WISE WEIGHTAGE (Term ‐ II)
No. Units Marks
II Algebra (Contd.) 23
III Geometry (Contd.) 17
IV Trigonometry (Contd.) 08
V Probability 08
VI Coordinate Geometry 11
VII Mensuration 23
Total 90
SYLLABUS
Unit II: Algebra (Contd.) 3. Quadratic Equations: Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solution of the quadratic equations
(only real roots) by factorization, by completing the square and by using quadratic formula. Relationship discriminant and nature of roots.
Problems related to day to day activities to be incorporated. 4. Arithmetic Progressions: Motivation for studying Arithmetic Progression, Derivation of the nth term and sum of first n terms of
A.P. and their application in solving daily life problems. Unit III: Geometry (Contd.) 2. Circles: Tangents to a circle motivated by chords drawn from points coming closer and closer to the point. i. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of
contact. ii. (Prove) The lengths of tangents drawn from an external point to circle are equal. 3. Constructions: i. Division of a line segment in a given ratio (internally). ii. Tangent to a circle from a point outside it. iii. Construction of a triangle similar to a given triangle. Unit IV: Trigonometry (Contd.) 3. Heights and Distances: Simple problems on heights and distances. Problems should not involve more than two right triangles.
Angles of elevation / depression should be only 30, 45, 60. Unit V: Statistics and Probability 2. Probability: Classical definition of probability. Connection with probability as given in Class IX. Simple problems
on single events not using set notation. Unit VI: Coordinate Geometry 1. Lines (In two dimensions): Review the concepts of coordinate geometry done earlier including graphs of linear equations.
Awareness of geometraical representation of quadratic polynomials. Distance between two points and section formula (internal). Area of a triangle.
Unit VII: Mensuration 1. Areas Related to Circles: Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and
perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60, 90 and 120 only. Plane figures involving triangles, simple quadrilaterals and circle should be taken).
2. Surface areas and Volumes: i. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres,
hemispheres and right circular cylinders / cones. Frustum of a cone. ii. Problems involving converting one type of metallic solid into another and other mixed problems.
(Problems with combination of not more than two different solids be taken).
QUESTION PAPER
PATTER
N
Mat
hem
atic
s C
ode
No.
041
Tim
e :
3 H
our
M
ark
s :
90
(%)
Wei
ghta
ge
26%
26%
24%
16%
8%
100%
*On
e of
th
e L
A (
4 m
ark
s) w
ill b
e to
ass
ess
the
valu
es in
her
ent
in t
he
text
s.
Tot
al
Mar
ks
23
23
22
14
08
90
Lon
g A
nsw
er
(LA
)
(4 M
ark
s)
3 4 2 —
2*
11
4 =
44
Sh
ort
An
swer
-II
(SA
)
(3 M
ark
s)
2 1 3 4 —
10
3 =
30
Sh
ort
An
swer
-I
(SA
)
(2
Mar
ks)
2 1 2 1 —
6
2 =
12
Ver
y S
hor
t A
nsw
er
(VS
A)
(1 m
ark
)
1 2 1 —
—
4
1 =
4
Typ
olog
y of
Qu
esti
ons
Rem
emb
erin
g –
(Kn
owle
dge
bas
ed S
impl
e re
call
qu
esti
ons,
to
know
spe
cifi
c fa
cts,
ter
ms,
con
cept
s,
prin
cipl
es
or
theo
ries
; Id
enti
fy,
defi
ne
or
reci
te,
info
rmat
ion)
Un
der
stan
din
g –
(Com
pre
hen
sion
– t
o be
fam
ilia
r w
ith
mea
ning
an
d to
un
ders
tand
co
ncep
tual
ly,
inte
rpre
t, co
mpa
re,
cont
rast
, ex
plai
n, p
arap
hras
e or
in
terp
ret i
nfor
mat
ion)
Ap
pli
cati
on –
(U
se a
bstr
act
info
rmat
ion
in c
oncr
ete
situ
atio
n, t
o ap
ply
know
ledg
e to
new
sit
uati
ons;
Use
gi
ven
cont
ent
to
inte
rpre
t a
situ
atio
n,
prov
ide
an
exam
ple
or s
olve
a p
robl
em)
Hig
h
Ord
er
Th
ink
ing
skil
ls
– (A
nal
ysis
an
d
Syn
thes
is
– C
lass
ify,
co
mpa
re,
cont
rast
or
di
ffer
enti
ate
betw
een
diff
eren
t pi
eces
of
info
rmat
ion;
O
rgan
ize
and/
or
inte
grat
e un
ique
pi
eces
of
in
form
atio
n fr
om a
var
iety
of
sour
ces)
Cre
atin
g,
Eva
luat
ion
and
M
ult
i-D
isci
pli
nar
y (G
ener
atin
g ne
w i
deas
, pr
oduc
t or
way
s of
vie
win
g th
ings
. A
ppra
ise,
jud
ge,
and/
or j
usti
fy t
he v
alue
or
wor
th
of
a de
cisi
on
or
outc
ome
or
to
pred
ict
outc
omes
bas
ed o
n va
lues
)
TO
TA
L
No.
1.
2.
3.
4.
5.
Contents
NCERT Textbook
Chapter No. Chapter Name Page No.
04 Quadratic Equations
1
05 Arithmetic Progressions
50
07 Coordinate Geometry
105
09 Some Applications of Trigonometry
153
10 Circles
184
11 Constructions
216
12 Areas Related to Circles
262
13 Surface Areas and Volumes
320
15 Probability
369
Model Question Paper – I
394
Model Question Paper – II
397
1
Chapter 04: Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0 where a, b, c are real numbers, a 0. (It is called standard form of a quadratic equation). Examples: 4x2 + 2x + 1 = 0, x2 + 4 = 0, 6 – 4x x2 = 0 are quadratic equations. Things to Remember 1. Check whether the following are quadratic
equations: i. (x + 1)2 = 2(x – 3) ii. x2 – 2x = (–2)(3 – x) iii. (x – 2) (x + 1) = (x – 1) (x + 3) iv. (x – 3) (2x + 1) = x (x + 5) v. (2x – 1) (x – 3) = (x + 5) (x – 1) vi. x2 + 3x + 1 = (x – 2)2 vii. (x + 2)3 = 2x (x2 – 1) viii. x3 – 4x2 – x + 1 = (x – 2)3 Solution: i. The given equation is (x + 1)2 = 2(x 3) x2 + 2x + 1 = 2x 6 x2 + 7 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. Alternate Method: LHS = (x + 1)2 = x2 + 2x + 1 RHS = 2(x 3) = 2x 6 (x + 1)2 = 2(x 3) can be written as x2 + 2x + 1 = 2x 6 x2 + 7 = 0 It is of the form ax2 + bx + c = 0 the given equation is a quadratic equation. ii. The given equation is x2 – 2x = (–2)(3 – x) x2 – 2x = 6 + 2x
x2 4x + 6 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. iii. The given equation is (x – 2) (x + 1) = (x – 1) (x + 3) x2 + x – 2x – 2 = x2 + 3x – x 3 x2 – x – 2 = x2 + 2x 3 3x – 1 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation.
(It is a linear equation). iv. The given equation is (x – 3) (2x + 1) = x (x + 5) 2x2 + x – 6x – 3 = x2 + 5x x2 – 10x – 3 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. v. The given equation is (2x – 1) (x – 3) = (x + 5) (x – 1) 2x2 – 6x – x + 3 = x2 – x + 5x – 5 2x2 – 7x + 3 = x2 + 4x – 5 x2 – 11x + 8 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. vi. The given equation is x2 + 3x + 1 = (x – 2)2 x2 + 3x + 1 = x2 – 4x + 4 7x – 3 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation. vii. The given equation is (x + 2)3 = 2x (x2 – 1) x3 + 6x2 + 12x + 8 = 2x3 – 2x x3 – 6x2 – 14x – 8 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation. viii. The given equation is x3 – 4x2 – x + 1 = (x – 2)3 x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8 2x2 – 13x + 9 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. 2. Represent the following situations in the
form of quadratic equations: i. The area of a rectangular plot is
528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
04. Quadratic Equations
NCERT Exercise 4.1
Things to Remember
Any equation of the form p(x) = 0 where p(x)is a quadratic polynomial is a quadraticequation.
ax2 + bx + c = 0, a 0 (when terms are
written in decreasing order of their degrees)is a standard form of the equation.
Quadratic Equations
2
Class X: Mathematics
2
ii. The product of two consecutive positive integers is 306. We need to find the integers.
iii. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
iv. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution: i. Let the breadth of the rectangle be x m. the length of the rectangle = (2x + 1) m Area of rectangular plot = length breadth = (2x + 1) x m2 = (2x2 + x) m2 According to the given condition, 2x2 + x = 528 2x2 + x – 528 = 0 the required quadratic equation is
2x2 + x – 528 = 0 ii. Let the two consecutive integers be x and x + 1. Product of the two integers = x(x + 1) According to the given condition, x(x + 1) = 306 x2 + x = 306 x2 + x – 306 = 0 the required quadratic equation is x2 + x – 306 = 0 iii. Let Rohan’s present age be x years. Present age of Rohan’s mother = (x + 26) years After 3 years, Age of Rohan = (x + 3) years Age of Rohan’s mother = (x + 26 + 3) years = (x + 29) years According to the given condition, (x + 3) (x + 29) = 360 x2 + 29x + 3x + 87 = 360 x2 + 32x – 273 = 0 the required quadratic equation is x2 – 32x – 273 = 0 iv. Let the uniform speed of the train be x km/hr. Time taken by the train to cover a distance of
480 km = 480
xhrs ....
distance time =
speed
If the speed had been 8 km/h less, Then, speed of train = (x – 8) km/h Time taken by the train to cover a distance of
480 km = 480
8x hrs
According to the given condition,
480
8x – 480
x = 3
480 1 1
8x x
= 3
1 1
8x x
= 3
480
8
8
x x
x x
= 1
160
8
8x x = 1
160
x(x – 8) = (160) (8) x2 – 8x = 1280 x2 – 8x – 1280 = 0 the required quadratic equation is x2 – 8x – 1280 = 0 1. Which of the following are quadratic
equations? i. 3x2 7x + 5 = 0 ii. 3x3 – 8x2 + 8 = 5x + 5 iii. 10x = 4x2
iv. x + 1
x= 3x
v. (x – 4)2 + 1 = 5x – 3 vi. x(x + 2) + 9 = (x + 4) (x – 4) Solution: i. The given equation is 3x2 – 7x + 5 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. ii. The given equation is 3x3 – 8x2 + 8 = 5x + 5 3x3 – 8x2 – 5x + 3 = 0 It is not of the form ax2 + bx + c = 0 the given equation is not a quadratic equation. iii. The given equation is 10x = 4x2 4x2 – 10x = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. iv. The given equation is
x + 1
x= 3x
2 1x
x
= 3x
x2 + 1 = 3x2 2x2 – 1 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation.
Problems based on Exercise 4.1
3
Chapter 04: Quadratic Equations
v. The given equation is (x – 4)2 + 1 = 5x – 3 x2 – 8x + 16 + 1 = 5x – 3 x2 – 8x + 17 = 5x – 3 x2 – 13x + 20 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. vi. The given equation is x(x + 2) + 9 = (x + 4) (x – 4) x2 + 2x + 9 = x2 – 16 2x + 25 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation. 2. In a cricket match Harbhajan took three
wickets less than twice the number of wickets taken by Zahir. The product of the number of wickets taken by these two is 20. Represent the above situation in the form of quadratic equation. [CBSE 2015]
Solution: Let the number of wickets taken by Zahir be x. Number of wickets taken by Harbhajan = 2x 3. According to the given condition, x(2x – 3) = 20 2x2 – 3x = 20 2x2 – 3x – 20 = 0 the required quadratic equation is 2x2 – 3x – 20 = 0. 3. The sum of two numbers is 27. We need to
find the numbers if sum of their reciprocal
is 3
4. Represent the above situation in the
form of quadratic equation. Solution: Let the first number be x second number = 27 – x.
Reciprocal of first number = 1
x
Reciprocal of second number = 1
27 x
According to the given condition,
1 1
27x x
= 3
4
27
27
x x
x x
= 3
4
27
27x x= 3
4
9
27x x= 1
4
36 = x(27 – x) 36 = 27x – x2 x2 – 27x + 36 = 0 the required quadratic equation is x2 – 27x + 36 = 0.
1. Check which of the following are quadratic
equations:
i. (x – 3)2 + 1 = 4x – 3
ii. x (2x + 4) = (x + 5) (x – 5)
iii. (x + 3)3 = x3 – 10
iv. x (4x + 5) = 4x2 + 5
v. (5x + 1) (4x + 2) = 20 (x – 1) (x – 2)
vi. 2x2 = 11
vii. 2
1x
x
= 12 7x
x
viii. 4x2 + 3 x + 9 = 3 x + 9 2. Represent the following situations in the
form of quadratic equations:
i. The product of two consecutive positive integers is 342. Represent this in the form of quadratic equation whose roots are these integers.
ii. The sum of the areas of two squares is 802 m2. The difference of their perimeter is 8 m. Represent a quadratic equation to find the sides of the two squares.
iii. Three consecutive even integers are such that the sum of the square of the first and the product of the other two is 484. Represent the quadratic equation to find out the integers.
iv. Sanika wishes to arrange three sticks together in the shape of a right triangle. The hypotenuse is 1 cm longer than the base and 8 cm longer than the altitude of the triangle. Form a quadratic equation to find the length of the smallest rod.
1. (i), (ii), (iii), (vi), (viii) are quadratic equations.
2. i. x2 + x – 342 = 0, where x is the smaller integer.
ii. x2 + 2x – 399 = 0, where x is side of smaller square.
iii. x2 + 3x – 238 = 0, where x is the smaller integer.
iv. x2 – 2x 15 = 0, where x is the length of smallest stick.
Practice Problems based on Exercise 4.1
Answers
4
Class X: Mathematics
4
1. Which of the following is a quadratic
equation? [NCERT Exemplar] (A) x2 + 2x + 1 = (4 – x)2 + 3
(B) 2x2 = (5 – x) 22
5x
(C) (k + 1) x2 + 3
2x = 7 where k = –1
(D) x3 – x2 = (x – 1)3 2. Which of the following is not a quadratic
equation? [NCERT Exemplar] (A) 2(x – 1)2 = 4x2 – 2x + 1 (B) 2x – x2 = x2 + 5
(C) 22 3x = 3x2 – 5x
(D) (x2 + 2x)2 = x4 + 3 + 4x2 3. Is 5 x2 + 6x + 3 = 0 a quadratic equation?
[CBSE 2012] (A) Yes (B) No (C) Can’t say (D) It is a linear equation 4. The standard form of quadratic equation is (A) ax3 + bx2 + cx + d = 0, a 0 (B) ax4 + cx + e = 0, a 0 (C) ax + b = 0, a 0 (D) ax2 + bx + c = 0, a 0 5. Which of the following is not a quadratic
equation? (A) (x – 5)2 + 1 = 4x – 3 (B) 3x (3x + 1) + 8 = 9(x + 2) (x – 2) (C) x (2x + 3) = x2 + 1 (D) (x + 2)3 = x3 – 4 6. Which of the following is a quadratic
equation? (A) (x + 1)2 = 5(x – 9) (B) (x – 6) (x + 7) = (x – 4) (x + 3) (C) x2 + 3x + 1 = (x – 2)2 (D) x3 – 1 = 0 In general, a real number is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a2 + b + c = 0. (where x = is solution of the quadratic equation). Factorisation Method: If the quadratic polynomial ax2 + bx + c can be expressed as the product of two linear factors, say (px + q) and (rx + s), where p, q, r, s R, such that p ≠ 0 and r ≠ 0,
then ax2 + bx + c = 0 can be written as
(px + q) (rx + s) = 0
px + q = 0 or rx + s = 0
q
p x or s
r x
Thus, q
p
and
s
r
are the roots of the quadratic
equation. Example:
x2 + 5x + 6 = 0
x2 + 3x + 2x + 6 = 0
x(x + 3) + 2(x + 3) = 0
(x + 3) (x + 2) = 0
x + 3 = 0 or x + 2 = 0
x = 3 or x = 2
x = –3 and x = –2 are two roots of the equation
x2 + 5x + 6 = 0. Note: The equation of the form (x – )2 = 0 has
repeated roots i.e. x = and x = 1. Find the roots of the following quadratic
equations by factorization:
i. x2 – 3x – 10 = 0
ii. 2x2 + x – 6 = 0
iii. 22 7 5 2 0 x x
iv. 2 12 + = 0
8x x
v. 100x2 – 20x + 1 = 0
Solution:
i. x2 – 3x – 10 = 0
x2 – 5x + 2x – 10 = 0
x (x – 5) + 2(x 5) = 0
(x – 5) (x + 2) = 0
x – 5 = 0 or x + 2 = 0
x = 5 or x = 2
the roots of the given equation are 2, 5.
Multiple Choice Questions
NCERT Exercise 4.2
Solution of a Quadratic Equation by Factorisation
Roots of the quadratic equationax2 + bx + c = 0 are the same as the zeroesof the quadratic polynomial ax2 + bx + c.
A quadratic equation can have atmost tworoots.
Things to Remember
5
Chapter 04: Quadratic Equations
ii. 2x2 + x – 6 = 0 2x2 + 4x – 3x – 6 = 0 2x (x + 2) – 3(x + 2) = 0 (2x – 3) (x + 2) = 0 2x – 3 = 0 or x + 2 = 0
x = 3
2 or x = 2
the roots of the given equation are 2, 3
2.
iii. 22 7 5 2 0 x x + 22 2 5 + 5 2 0 x x + x
2 2 5 2 0x x+ x+
2 5 2 0x+ x+
2 5 = 0x + or + 2 0x
5
=2
x or 2x =
the roots of the given equation are 5
2
, 2 .
iv. 2 1
28
x x = 0
16x2 – 8x + 1 = 0 16x2 – 4x – 4x + 1 = 0 4x (4x – 1) – 1(4x – 1) = x0 (4x – 1) (4x – 1) = 0 4x – 1 = 0 or 4x – 1 = 0
1 =
4x or 1
= 4
x
the roots of the given equation are 1
4, 1
4.
v. 100x2 – 20x + 1 = 0 100x2 – 10x – 10x + 1 = 0 10x (10x – 1) – 1(10x – 1) = 0 (10x – 1) (10x – 1) = 0 10x – 1 = 0 or 10x – 1 = 0
1=
10x or 1
=10
x
the roots of the given equation are 1 1,
10 10.
2. Represent the following situations
mathematically, and solve them. i. John and Jivanti together have 45 marbles.
Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
ii. A cottage industry produces a certain
number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the
total cost of production was ` 750. We would like to find out the number of toys produced on that day.
Solution: i. Let the number of marbles with John be x. Number of marbles with Jivanti = 45 – x. Number of marbles left with John when he
lost 5 marbles = x – 5 Number of marbles left with Jivanti when she
lost 5 marbles = 45 – x – 5 = 40 x According to the given condition, (x – 5) (40 – x) = 124 40x – x2 – 200 + 5x = 124 – x2 + 45x – 200 = 124 – x2 + 45x – 324 = 0 x2 – 45x + 324 = 0 x2 – 36x – 9x + 324 = 0 x(x – 36) – 9(x – 36) = 0 (x – 36) (x – 9) = 0 x – 36 = 0 or x – 9 = 0 x = 36 or x = 9 Number of marbles with Jivanti = 45 – x = 45 – 36 or 45 – 9 = 9 or 36 the number of marbles John and Jivanti had to
start with are either 36, 9 or 9, 36 respectively. ii. Let the number of toys produced in a day be x. Cost of production of each toy = ` (55 – x) The total cost of production on a particular
day = ` 750 According to the given condition, x(55 – x) = 750 55x – x2 = 750 x2 – 55x + 750 = 0 x2 – 30x – 25x + 750 = 0 x(x – 30) – 25(x – 30) = 0 (x – 30) (x – 25) = 0 x – 30 = 0 or x – 25 = 0 x = 30 or x = 25 the number of toys produced on that day are 25
or 30. 3. Find two numbers whose sum is 27 and
product is 182. Solution: Let one number be x. the other number = 27 – x According to the given condition, x(27 – x) = 182 27x – x2 = 182 x2 – 27x + 182 = 0 x2 – 14x – 13x + 182 = 0 x(x – 14) – 13(x – 14) = 0
6
Class X: Mathematics
6
(x – 14) (x – 13) = 0 x – 14 = 0 or x – 13 = 0 x = 14 or x = 13 other number = 27 x = 27 14 or 27 13 = 13 or 14 the two numbers are 13 and 14. 4. Find two consecutive positive integers, sum
of whose squares is 365. Solution: Let the two consecutive numbers be x and
x + 1. According to the given condition, x2 + (x + 1)2 = 365 x2 + x2 + 2x + 1 = 365 2x2 + 2x – 364 = 0 x2 + x – 182 = 0 x2 + 14x – 13x – 182 = 0 x(x + 14) – 13(x + 14) = 0 (x – 13) (x + 14) = 0 x – 13 = 0 or x + 14 = 0 x = 13 or x = 14 Since, x is positive integer x = 13 other number = x + 1 = 13 + 1 = 14 the two consecutive numbers are 13 and 14. 5. The altitude of a right triangle is 7 cm less
than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution: Let the base of the triangle be x cm. Altitude of the triangle = (x – 7) cm. By Pythagoras theorem, BC2 + AB2 = AC2 x2 + (x – 7)2 = (13)2 x2 + x2 – 14x + 49 = 169 2x2 – 14x – 120 = 0 x2 – 7x – 60 = 0 x2 – 12x + 5x – 60 = 0 x(x – 12) + 5(x – 12) = 0 (x – 12) (x + 5) = 0 x – 12 = 0 or x + 5 = 0 x = 12 or x = 5 But length of the side cannot be negative. x = 12 cm. Altitude of the triangle = x – 7 = 12 – 7 = 5 cm the length of other two sides are 5 cm and
12 cm.
6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article.
Solution: Let the number of articles produced be x. Cost of each article = ` (2x + 3) Total cost of production = x(2x + 3) According to the given condition, x(2x + 3) = 90 2x2 + 3x = 90 2x2 + 3x – 90 = 0 2x2 + 15x – 12x – 90 = 0 x(2x + 15) – 6(2x + 15) = 0 (x – 6) (2x + 15) = 0 x – 6 = 0 or 2x + 15 = 0
x = 6 or 15=
2
x
But the number of articles cannot be a negative fraction.
x = 6 Cost of each article = 2x + 3 = 2 6 + 3 = ` 15 the number of articles produced is 6 and the
cost of each article is ` 15. 1. If x =
1
2, is a solution of the quadratic
equation 3x2 + 2kx – 3 = 0, find the value of k. [CBSE 2015]
Solution: The given equation is 3x2 + 2kx – 3 = 0
Putting x = 1
2 , we get
2
1 13 2k 3 0
2 2
3k 3 0
4
3k 3
4
3 12k
4
9k
4
the value of k is 9
4
.
Problems based on Exercise 4.2 A
B C x
x – 713
7
Chapter 04: Quadratic Equations
2. Find the roots of the following quadratic equations by factorization:
i. x x2 3 +1 + 3 = 0 [CBSE 2015]
ii. 23 2 3x x = 0 [CBSE 2011, 2012]
iii. 24 3 + 5 2 3 = 0x x
[CBSE 2012, 2013] iv. abx2 + (b2 ac)x bc = 0 v. 9x2 – 6b2 x – (a4 – b4) = 0
[CBSE 2015] vi. (x – 1)2 – 5(x – 1) – 6 = 0 [CBSE 2015] vii. 2 9x + x = 13 Solution:
i. 2 3 1 + 3 0 x x
2 3 3 0 x x x +
3 1 3 0 x x x
3 1 0 x x
3 0 x or x – 1 = 0
= 3x or x = 1
the roots of the given equation are 3 , 1. ii. 23 2 3 0 x x
23 3 + 3 0 x x x
3 3 1 3 0 x x x
3 3 1 0x x
3 0 x or 3 +1 = 0x
= 3x or 1
=3
x
the roots of the given equation are 1
3,3
.
iii. 24 3 5 2 3 0 x x
24 3 8 3 2 3 0 x x x
4 3 + 2 3 3 + 2 0 x x x
4 3 3 + 2 0 x x
4 3 0 x or 3 + 2 = 0x
3=
2x or
2=
3x
the roots of the given equation are 3 2
,2 3
.
iv. abx2 + (b2 ac) x bc = 0 abx2 + b2x acx bc = 0 bx (ax + b) c (ax + b) = 0 (ax + b) (bx c) = 0
ax + b = 0 or bx c = 0
x = b
a
or x =
c
b
the roots of the given equation are b
a
,
c
b.
v. 9x2 – 6b2 x – (a4 – b4) = 0 9x2 – [3(a2 + b2) – 3(a2 – b2)] x
– (a2 + b2) (a2 – b2) = 0 9x2 – 3(a2 + b2) x + 3(a2 – b2) x
– (a2 + b2) (a2 – b2) = 0 3x [3x – (a2 + b2)] + (a2 – b2) [3x
– (a2 + b2)] = 0 [3x – (a2 + b2)] [3x + a2 – b2] = 0 3x – (a2 + b2) = 0 or 3x + a2 – b2 = 0 3x = a2 + b2 or 3x = b2 – a2
x = 2 2a b
3
or x =
2 2b a
3
the roots of the given equation are 2 2a b
3
,
2 2b a
3
.
vi. (x – 1)2 – 5(x – 1) – 6 = 0 x2 – 2x + 1 – 5x + 5 – 6 = 0 x2 – 7x = 0 x(x – 7) = 0 x = 0 or x – 7 = 0 x = 0 or x = 7 the roots of the given equation are 0, 7. vii. 2 9x + x = 13 2 9x = 13 – x 2x + 9 = (13 – x)2
….[Squaring both sides] 2x + 9 = 169 – 26x + x2 x2 – 28x + 160 = 0 x2 – 20x – 8x + 160 = 0 x(x 20) 8(x – 20) = 0 (x – 20) (x – 8) = 0 x 20 = 0 or x 8 = 0 x = 20 or x = 8 the roots of the given equation are 20, 8. 3. Find the roots of the following quadratic
equations:
i. 110 3, 0x x
x [CBSE 2012]
ii. 13 1 3
3
x x x
[CBSE 2011, 2012]
iii. 16 151 ; 0, 1
1x
x x
[CBSE 2014]
8
Class X: Mathematics
8
iv. 1 1 6
, 1, 51 5 7
x
x xfor x:
[CBSE 2010]
v.
2 1 3 9 30; 3,
3 2 3 3 2 3 2
x xx
x x x x
[CBSE 2012]
vi. 2 1 3 12 3 5; 3,
3 2 1 2
x xx
x x
[CBSE 2014]
vii. 1 1 1 1= + +
a + b + a bx x [CBSE 2012]
Solution:
i. 110 3 x
x
10x2 – 1 = 3x
10x2 – 3x – 1 = 0
10x2 5x + 2x – 1 = 0
5x (2x 1) + 1(2x 1) = 0
(5x + 1) (2x 1) = 0
5x + 1 = 0 or 2x 1 = 0
1
5
x or 1
2x
the roots of the given equation are
1 1,
5 2 .
ii. 13 1 3
3
x x x
x2 + 2x – 3 = 3x – 1
x2 – x – 2 = 0
x2 – 2x + x – 2 = 0
x(x – 2) + 1(x – 2) = 0
(x – 2) (x + 1) = 0
x – 2 = 0 or x + 1 = 0
x = 2 or x = –1
the roots of the given equation are 2, –1.
iii. 16 151
1
x x
16 15
1
x
x x
(16 – x) (x + 1) = 15x
16x + 16 – x2 – x = 15x
– x2 + 15x + 16 = 15x
x2 = 16
x = 16
x = 4
the roots of the given equation are 4, – 4.
iv. 1 1 6
1 5 7
x x
5 1 6
1 5 7
x x
x x
2
5 1 6
5 5 7
x x
x x x
2
6 6
4 5 7
x x
2
1 1
4 5 7x x
x2 + 4x 5 = 7 x2 + 4x 12 = 0 x2 + 6x 2x 12 = 0 x (x + 6) 2 (x + 6) = 0 (x + 6) (x 2) = 0 x + 6 = 0 or x 2 = 0 x = 6 or x = 2 the roots of the given equation are –6, 2.
v.
2 1 3 90
3 2 3 3 2 3
x x
x x x x
2x(2x + 3) + (x – 3) + (3x + 9) = 0 4x2 + 6x + x – 3 + 3x + 9 = 0 4x2 + 10x + 6 = 0 2x2 + 5x + 3 = 0 2x2 + 2x + 3x + 3 = 0 2x(x + 1) + 3(x + 1) = 0 (2x + 3) (x + 1) = 0 2x + 3 = 0 or x + 1 = 0
3
2 x or x = –1
But, x ≠ 3
2
x = –1 the root of the given equation is –1.
vi. 2 1 32 3 5
3 2 1
x x
x x
4 2 3 95
3 2 1
x x +
x x
4 2 2 1 3 9 35
3 2 1
x x x x
x x
(4x – 2) (2x – 1) – (3x + 9) (x + 3) = 5(x + 3) (2x – 1)
(8x2 – 4x – 4x + 2) – (3x2 + 9x + 9x + 27) = 5(2x2 – x + 6x – 3)
(8x2 – 8x + 2) – (3x2 +18x + 27) = 5(2x2 + 5x – 3)
5x2 – 26x – 25 = 10x2 + 25x – 15 5x2 + 51x + 10 = 0 5x2 + 50x + x + 10 = 0
9
Chapter 04: Quadratic Equations
5x(x + 10) + 1(x + 10) = 0 (5x + 1) (x + 10) = 0 5x + 1 = 0 or x + 10 = 0
1
5 x or x = –10
the roots of the given equation are 1
5
, –10.
vii. 1 1 1 1
a b a b
x x
1 1 1 1
a b a b
x x
a b a b
a b ab
x x
x x
a b a b
a b ab
x x
1 1
a b abx x
– ab = x(a + b + x) xa + xb + x2 + ab = 0 x2 + (a + b)x + ab = 0 x2 + ax + bx + ab = 0 x (x + a) + b (x + a) = 0 (x + a) (x + b) = 0 x + a = 0 or x + b = 0 x = – a or x = – b the roots of the given equation are –a, –b. 4. If (x2 + y2) (a2 + b2) = (ax + by)2. Prove that
a b
x y [CBSE 2014]
Solution: (x2 + y2) (a2 + b2) = (ax + by)2 x2a2 + x2b2 + y2a2 + y2b2
= a2x2 + b2y2 + 2abxy x2b2 + y2a2 – 2abxy = 0 (xb – ya)2 = 0 xb = ya
a b
x y
Hence proved. 5. The sum of ages (in years) of a son and his
father is 35 years and product of their ages is 150 years, find their ages.
[CBSE 2012, 2014] Solution: Let the age of son be x years Age of father = (35 – x) years According to the given condition, x(35 – x) = 150 35x – x2 = 150 x2 – 35x + 150 = 0
x2 – 30x – 5x + 150 = 0 x(x – 30) – 5(x – 30) = 0 (x – 30) (x – 5) = 0 x – 30 = 0 or x 5 = 0 x = 30 or x = 5 But age of son cannot be 30 years, because
that would mean father’s age = 5 years which is not possible.
x = 5 years Age of father = (35 – x) = 35 – 5 = 30 years the ages of son and his father are 5 years and
30 years respectively. 6. Find the two consecutive odd positive integers,
sum of whose square is 290. [CBSE 2012] Solution: Let the two consecutive odd positive integers
be x and x + 2. According to the given condition, x2 + (x + 2)2 = 290 x2 + x2 + 4x + 4 = 290 2x2 + 4x – 286 = 0 x2 + 2x – 143 = 0 x2 + 13x – 11x – 143 = 0 x(x + 13) – 11(x + 13) = 0 (x + 13) (x – 11) = 0 x + 13 = 0 or x – 11 = 0 x = –13 or x = 11 Since, x is positive integer, x = 11 other number = x + 2 = 11 + 2 = 13 the two consecutive odd positive integers are
11 and 13. 7. The difference of two numbers is 5 and the
difference of their reciprocals is 1
10. Find
the numbers. [CBSE 2012, 2014] Solution: Let one number be x Other number = x + 5 According to the given condition,
1 1 1
5 10
x x
5 1
5 10
x x
x x
5 1
5 10
x x
50 = x(x + 5) x2 + 5x – 50 = 0 x2 + 10x – 5x – 50 = 0 x(x + 10) – 5(x + 10) = 0 (x + 10) (x – 5) = 0
10
Class X: Mathematics
10
x + 10 = 0 or x – 5 = 0 x = – 10 or x = 5 Other number = x + 5 = 10 + 5 or 5 + 5 = 5 or 10 the two numbers are 5, 10 or 10, 5. 8. The total cost of a certain length of cloth is `
200. If the piece was 5 m longer and each metre of cloth costs ` 2 less, the cost of the piece would have remained unchanged. How longer is the piece and what is its original rate per metre? [CBSE 2015]
Solution: Let the length of the cloth be x m. Total cost of the cloth = ` 200
Cost per metre = ` 200
x
New length of the cloth = (x + 5) m
New cost per metre = ` 200
2 x
Total cost of the cloth = (x + 5) 2002
x
According to the given condition,
2005 2 200
xx
1000200 2 10 200 x
x
– 2x + 1000
x– 10 = 0
– 2x2 + 1000 – 10x = 0 x2 + 5x – 500 = 0 x2 + 25x – 20x – 500 = 0 x(x + 25) – 20(x + 25) = 0 (x + 25) (x – 20) = 0 x + 25 = 0 or x – 20 = 0 x = – 25 or x = 20 Length of cloth = 20 m
Original cost per metre = 200
x = ` 200
20= ` 10
the length of cloth is 20 m and its original rate is ` 10 per metre.
9. A shopkeeper buys a number of books for `1200. If he had bought 10 more books for the same amount, each book would have cost him `20 less. How many books did he buy? [CBSE 2012]
Solution: Let the number of books bought be x. Total cost of books = ` 1200
Cost of each book = ` 1200
x
New number of books = x + 10
Cost of each book = ` 1200
10x
According to the given condition,
1200 120020
10
x x
1200 10 120020
10
x x
x x
1200x + 12000 – 1200x = 20x(x + 10) 12000 = 20x2 + 200x 600 = x2 + 10 x x2 + 10x – 600 = 0 x2 + 30x – 20x – 600 = 0 x(x + 30) – 20(x + 30) = 0 (x + 30) (x – 20) = 0 x + 30 = 0 or x – 20 = 0 x = – 30 or x = 20 But, number of books cannot be negative. x = 20 the number of book bought is 20. 10. A takes 6 days less than the time taken by B
to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.
[CBSE 2012] Solution: Let B finish the work in x days. Time take by A to finish the work = (x 6) days
Part of work done by A in 1 day = 1
6x
Part of work done by B in 1 day = 1
x
Time taken by A and B to finish the work together = 4 days
Part of work done by A and B in 1 day = 1
4
According to the given condition,
1 1 1
6 4
x x
6 1
6 4
x x
x x
2 6 1
6 4
x
x x
4(2x – 6) = x(x – 6) 8x – 24 = x2 – 6x x2 – 14x + 24 = 0 x2 – 12x – 2x + 24 = 0 x(x – 12) – 2(x – 12) = 0 (x – 12) (x – 2) = 0 x – 12 = 0 or x – 2 = 0 x = 12 or x = 2 But x cannot be less than 6 x = 12 the time taken by B to finish the work is 12 days.
11
Chapter 04: Quadratic Equations
11. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original
fraction is 29
20. Find the original fraction.
[CBSE 2015] Solution: Let the denominator of the fraction be x Numerator of the fraction = x – 3
Original Fraction = 3x
x
New fraction = 3 2
2
x
x
1
2
x
x
According to the given condition,
3 1 29
2 20
x x
x x
2 3 1 29
2 20
x x x x
x x
2 2
2
3 2 6 29
2 20
x x x x x
x x
2
2
2 2 6 29
2 20
x x
x x
20(2x2 – 2x – 6) = 29(x2 + 2x) 40x2 – 40x – 120 = 29x2 + 58x 11x2 – 98x – 120 = 0 11x2 – 110x + 12x – 120 = 0 11x(x – 10) + 12(x – 10) = 0 (11x + 12) (x – 10) = 0 11x + 12 = 0 or x – 10 = 0
12=
11x or x = 10
x = 12
11
does not satisfy the given condition
x = 10
Original fraction = 3x
x
= 10 3
10
= 7
10
the original fraction is 7
10.
12. A motor boat, whose speed is 15 km/h in
still water, goes 30 km downstream and comes back in a total of 4 hours and 30 minutes. Determine the speed of streams.
[CBSE 2012, 2014] Solution: Let the speed of the stream = x km/h. Speed of motorboat in upstream = (15 – x) km/h Speed of motor boat in downstream = (15 + x) km/h
Downstream distance = Upstream distance = 30 km
Time required to go downstream = 30
15 x hrs
Time required to go upstream = 30
15 x hrs
According to the given condition,
30 30 14
15 15 2
x x
30 30 9
15 15 2
x x
30 15 30 15 9
15 15 2
x x
x x
2
450 30 450 30 9
225 2
x x
x
2
900 9
225 2x
2
100 1
225 2x
200 = 225 – x2 x2 = 25 x = 25 x = 5 But, speed cannot be negative. x = 5 the speed of the stream is 5 km/h. 13. A peacock is sitting on top of a pillar which
is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake, the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught?
Solution: AB is the pillar. Peacock is at point A. Snake’s initial position is D. It is moving in
the direction of its hole which is at point B. Let the snake be caught at a distance x m from
the hole. Since the speed of the peacock and snake are
equal, they cover equal distance till point C. AC = DC = DB BC = (27 – x) m
A
9 m
B C Dx m
27 m
36
Class X: Mathematics
36
Since the roots of the equation are equal, b2 4ac = 0 (2p)2 4(p)(6) = 0 4p2 24p = 0 4p2 24p = 0 4p(p 6) = 0 4p = 0 or p 6 = 0 p = 0 or p = 6 But p ≠ 0 p = 6 the value of p is 6. v. (p 12)x2 + 2(p 12) x + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = p 12, b = 2(p 12), c = 2 Since the roots of the equation are equal, b2 4ac = 0 [2(p 12)]2 4(p 12) 2 = 0 4(p 12)2 8(p 12) = 0 4 (p 12) (p 12 2) = 0 (p 12) (p 14) = 0 p 12 = 0 or p 14 = 0 p = 12 or p = 14 But p ≠ 12 p = 14 the value of p is 14. 2. Find the value of p for which the quadratic
equation (p + 1) x2 6(p + 1)x + 3(p + 9) = 0 p 1, has equal roots hence find the roots of the equation. [CBSE 2015]
Solution: (p + 1)x2 6(p + 1)x + [3(p + 9)] = 0 Comparing with ax2 + bx + c = 0, we get a = p + 1, b = 6(p + 1), c = 3(p + 9) Since the roots of the equation are equal, b2 4ac = 0 [6(p + 1)]2 4(p + 1) [3(p + 9)] = 0 36 (p + 1)2 12(p + 1) (p + 9) = 0 12 (p + 1) [3(p + 1) (p + 9)] = 0 (p + 1) (3p + 3 p 9) = 0 (p + 1) (2p 6) = 0 p + 1 = 0 or 2p 6 = 0 p = 1 or p = 3 But p –1 p = 3 Substituting the value of p in the given
equation, we get (3 + 1)x2 6 (3 + 1)x + 3 (3 + 9) = 0 4x2 24x + 36 = 0 x2 6x + 9 = 0 (x 3)2 = 0 x = 3 or x = 3 the roots of the given equation are 3, 3.
3. If 2 is a root of the quadratic equation 3x2 + px 8 = 0 and the quadratic equation 4x2 2px + k = 0 has equal roots, find k.
[CBSE 2014]
Solution:
2 is a root of the equation 3x2 + px 8 = 0
3(2)2 + p(2) 8 = 0
12 + 2p 8 = 0
2p + 4 = 0
2p = 4
p = 2
Substituting the value of p in 4x2 2px + k = 0, we get
4x2 2(2) x + k = 0
4x2 + 4x + k = 0
Comparing with ax2 + bx + c = 0, we get
a = 4, b = 4, c = k
Since the roots of the equation are equal,
b2 4ac = 0
42 4(4)(k) = 0
16 16k = 0
16k = 16
k = 1
the value of k is 1. 4. Find the nature of the roots of the following
equation. If the real roots exists, find them:
3x2 4 3 x + 4 = 0 [CBSE 2012]
Solution:
3x2 4 3 x + 4 = 0
Comparing with ax2 + bx + c = 0, we get
a = 3, b = 4 3 , c = 4
b2 4ac = 2
4 3 4(3)(4)
= 48 48
= 0
Since, b2 4ac = 0,
the roots are real and equal.
x = b b
and2a 2a
x =4 3 4 3
and2 3 2 3
x = 2 3 2 3
and3 3
the roots of the given equation are 2 3
3,
2 3
3.
37
Chapter 04: Quadratic Equations
5. Determine the positive value of ‘k’ for which the equation x2 + kx + 64 = 0 and x2 8x + k = 0 will both have real and equal roots. [CBSE 2014]
Solution: The first equation is x2 + kx + 64 = 0 Comparing with ax2 + bx + c = 0, we get Here a = 1, b = k, c = 64 Since the roots of the equation are equal, b2 4ac = 0 k2 4(1)(64) = 0 k2 = 256 k = 256 k = ± 16 ....(i) The second equation is x2 8x + k = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 8, c = k Since the roots of the equation are equal, b2 4ac = 0 (8)2 4(1)(k) = 0 64 4k = 0 4k = 64
k = 64
4
k = 16 ....(ii) From (i) and (ii), we get k = 16. the value of k is 16. 6. If the quadratic equation (1 + a2) b2x2 + 2abcx + c2 m2 = 0 in x has
equal roots prove that c2 = m2(1 + a2). [CBSE 2014, 2015]
Solution: (1 + a2) b2x2 + 2 abcx + (c2 m2) = 0 Comparing with Ax2 + Bx + C = 0, we get A = (1 + a2)b2, B = 2abc, C = c2 m2 Since the roots of the equation are equal, B2 4AC = 0 (2abc)2 4(1 + a2) b2(c2 m2) = 0 4a2b2c2 4(b2 + a2b2) (c2 m2) = 0 4a2b2c2 4[b2c2 b2m2 + a2b2c2 a2b2m2] = 0 4a2b2c2 4b2c2 + 4b2m2 4a2b2c2
+ 4a2b2m2 = 0 4b2 [a2m2 + m2 c2] = 0 a2m2 + m2 c2 = 0 c2 = a2m2 + m2 c2 = m2(1 + a2) Hence proved. 7. For what values of k will quadratic
equation (2k + 1)x2 + 2(k + 3)x + (k + 5) = 0 have real and equal roots? [CBSE 2012]
Solution: (2k + 1)x2 + 2(k + 3)x + (k + 5) = 0 Comparing with ax2 + bx + c = 0, we get a = 2k + 1, b = 2(k + 3), c = k + 5
Since the roots of the equation are equal, b2 4ac = 0 [2(k + 3)]2 4(2k + 1) (k + 5) = 0 4(k + 3)2 4(2k2 + 11k + 5) = 0 (k + 3)2 (2k2 + 11k + 5) = 0 k2 + 6k + 9 2k2 11k 5 = 0 k2 5k + 4 = 0 k2 + 5k 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 5, c = 4 By quadratic formula,
x = 2b b 4ac
2a
k =
25 5 4 1 4
2 1
k = 5 25 16
2
k = 5 41
2
k = 5 41
2
or k =
5 41
2
the given equation has real and equal roots
when the value of k is5 41 5 41
or .2 2
8. Find the nature of the roots of the
quadratic equation 13 3 x2 + 10x + 3 = 0 [CBSE 2012]
Solution: 13 3 x2 + 10x + 3 = 0 Comparing with ax2 + bx + c = 0, we get
a = 13 3 , b = 10, c = 3
b2 4ac = (10)2 4 13 3 3
= 100 156 = 56 < 0 Since b2 4ac < 0, the equation has no real roots. 9. Find whether the real roots of the quadratic
equation 2 x2 + 7x + 8 2 = 0 exist or not. [CBSE 2012]
Solution: The given equation is 2 x2 + 7x + 8 2 = 0 Comparing with ax2 + bx + c = 0, we get
a = 2 , b = 7, c = 8 2
b2 4ac = (7)2 4 2 8 2
= 49 64 = 15 < 0 Since, b2 4ac < 0, the equation has no real roots.
38
Class X: Mathematics
38
10. In px2 + 4 3 x + 3 = 0 find the value of p so that
i. the roots are real ii. the roots are not real iii. the roots are equal
[CBSE 2012] Solution: The given equation is px2 + 4 3 x + 3 = 0 Comparing with ax2 + bx + c = 0, we get
a = p, b = 4 3 , c = 3
b2 4ac = 2
4 3 4(p)(3) = 48 12p
i. If the roots are real, b2 4ac ≥ 0 48 12p ≥ 0 48 ≥ 12p 4 ≥ p p ≤ 4 ii. If the roots are not real, b2 4ac < 0 48 12p < 0 48 < 12p 4 < p p > 4 iii. If roots are equal, b2 4ac = 0 48 12p = 0 12p = 48 p = 4 11. Rina wishes to fit three rods together in a
shape of a right triangle. The hypotenuse is to be 1 cm longer than the base and 2 cm longer than the altitude. Is it possible to draw such type of a right triangle. If so what should be the lengths of the rods?
Solution: Let the length of the hypotenuse x cm. Length of the base = (x 1) cm Length of the altitude = (x 2) cm In right ABC, by Pythagoras theorem, AC2 = AB2 + BC2 x2 = (x 2)2 + (x 1)2 x2 = x2 4x + 4 + x2 2x + 1 x2 6x + 5 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 6, c = 5 b2 4ac = (6)2 4(1)(5) = 36 20 = 16 > 0 Since b2 4ac > 0, the roots are real and distinct.
Hence it is possible to can draw the triangle. The roots of the equation are given by
x = 2b b 4ac
2a
x =
6 16
2 1
x = 6 4
2
x = 6 4
2
or x =
6 4
2
x = 10
2 or x =
2
2
x = 5 or x = 1 But x ≠ 1 x = 5 cm AB = x 2 = 5 2 = 3 cm BC = x 1 = 5 1 = 4 cm the lengths of the three rods are 5 cm, 3 cm
and 4 cm. 1. State whether the following quadratic
equations have two distinct real roots. Justify your answer.
i. x2 – 3x + 4 = 0 ii. 2x2 + x – 1 = 0
iii. 2x2 – 6x +9
2 = 0
iv. 3x2 – 4x + 1 = 0 v. (x + 4)2 – 8x = 0
vi. (x – 2 )2– 2(x + 1) = 0
vii. 2 x2 3
2x +
1
2= 0
viii. x (1 – x) – 2 = 0 ix. (x – 1)(x + 2) + 2 = 0 x. (x + 1)(x – 2) + x = 0 Solution: i. No Justification: The given equation is x2 3x + 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 3, c = 4 b2 4ac = (3)2 4(1)(4) = 9 16 = 7 < 0 Since b2 4ac < 0, the roots of the given equation are not real.
NCERT Exemplar
x
(x 1)
(x 2)
A
B C
39
Chapter 04: Quadratic Equations
ii. Yes Justification: The given equation is 2x2 + x 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = 1, c = 1 b2 4ac = (1)2 4(2)(1) = 1 + 8 = 9 > 0 Since b2 4ac > 0, the roots of the given equation are real and
distinct. iii. No Justification:
The given equation is 2x2 6x + 9
2 = 0
4x2 12x + 9 = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = 12, c = 9 b2 4ac = (12)2 4(4)(9) = 144 144 = 0 Since, b2 4ac = 0 the roots of the given equation are real and
equal. iv. Yes Justification: The given equation is 3x2 4x + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 3, b = 4, c = 1 b2 4ac = (4)2 4(3)(1) = 16 12 = 4 > 0 Since, b2 4ac > 0 the roots of the given equation are real and
distinct. v. No Justification: The given equation is (x + 4)2 8x = 0 x2 + 8x + 16 8x = 0 x2 + 16 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 0, c = 16 b2 4ac = (0)2 4(1)(16) = 64 < 0 Since b2 4ac < 0, the roots of the given equation are not real. vi. Yes Justification: The given equation is (x 2 )2 2(x + 1) = 0
x2 2 2 x + 2 2x 2 = 0
x2 2 2 x 2x = 0
x2 (2 2 + 2)x = 0
Comparing with ax2 + bx + c = 0, we get
a = 1, b = 2 2 2 , c = 0
b2 4ac = [(2 2 + 2)]2 4(1)(0)
= 8 + 8 2 + 4 0
= 12 + 8 2 > 0 Since b2 4ac > 0, the roots of the given equation are real and
distinct. vii. Yes Justification:
The given equation is 2 x2 3
2 x +
1
2 = 0
2x2 3x + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = 3, c = 1 b2 4ac = ( 3)2 4(2)(1) = 9 8 = 1 > 0 Since b2 4ac > 0, the roots of the given equation are real and
distinct. viii. No Justification: The given equation is x(1 x) 2 = 0 x x2 2 = 0 x2 x + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 1, c = 2 b2 4ac = (1)2 4(1)(2) = 1 8 = 7 < 0 Since b2 4ac < 0, the roots of the given equation are not real. ix. Yes Justification: The given equation is (x 1)(x + 2) + 2 = 0 x2 + 2x x 2 + 2 = 0 x2 + x = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 1, c = 0 b2 4ac = 12 4(1)(0) = 1 > 0 Since b2 4ac > 0, the roots of the given equation are real and
distinct. x. Yes Justification: The given equation is (x + 1) (x 2) + x = 0 x2 x 2 + x = 0 x2 – 2 = 0
40
Class X: Mathematics
40
Comparing with ax2 + bx + c = 0, we get a = 1, b = 0, c = 2 b2 4ac = 02 4(1)(2) = 8 > 0 Since b2 4ac > 0, the roots of the given equation are real and
distinct. 2. Write whether the following statements are
true or false. Justify your answers. i. Every quadratic equation has exactly
one root. ii. Every quadratic equation has at least
one real root. iii. Every quadratic equation has at least
two roots. iv. Every quadratic equations has at
most two roots. v. If the coefficient of x2 and the
constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
vi. If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.
vii. If in a quadratic equation the coefficient of x is zero, then the quadratic equation has no real roots.
Solution: i. False. Justification: Every quadratic equation has two roots. ii. False. Justification: If the discriminant b2 4ac < 0, then the
quadratic equation has no real roots. iii. False. Justification: Every quadratic equation has only two roots. It
cannot have more than two roots. iv. True. Justification: A quadratic equation cannot have more than
two roots. v. True. Justification: If the values of a and c are of opposite sign,
then b2 4ac > 0. Hence the roots are real. vi. True. Justification: If the values of a and c are of same sign, and b
is 0, then b2 4ac < 0 Hence the roots are not real. vii. Not always.
Justification: Consider the quadratic equation ax2 + c = 0 Comparing with Ax2 + Bx + C = 0, we get A = a, B = 0, C = c B2 4AC = 4ac If a and c have same sign, then B2 4AC < 0 the roots are not real. If a and c have opposite signs, then
B2 4AC > 0 the roots are real and distinct. 3. A quadratic equation with integral
coefficient has integral roots. Justify your answer.
Solution: Not always. Justification: Consider x2 + x + 2 = 0 The equation has integral coefficients a = 1, b = 1, c = 2 b2 4ac = 12 4(1)(2) = 1 8 = 7 < 0 Since, b2 4ac < 0 the equation has no real roots. 4. Does there exist a quadratic equation whose
coefficients are rational but both of its roots are irrational? Justify your answer.
Solution: Yes Justification: Consider the equation x2 3x + 1 = 0 The equation has rational coefficients a = 1, b = 3, c = 1 Now, b2 4ac = (3)2 4(1) (1) = 9 4 = 5 > 0 the roots are given by
x =
3 5
2 1
= 3 5
2
the roots are irrational. 5. Does there exist a quadratic equation whose
coefficients are all distinct irrationals but both the roots are rationals? Why?
Solution Yes Justification: Consider the equation 5 x2 5 5 x + 6 5 =
0 25 3 5 2 5 6 5 0 x x x
5 3 2 5 3 0 x x x x
3 5 2 5 0 x x
41
Chapter 04: Quadratic Equations
x 3 = 0 or 5 2 5 0 x x = 3 or x = 2 the roots of the equation are 2 and 3, which are
rational. 6. Is 0.2 a root of the equation x2 – 0.4 = 0?
Justify. Solution: No Justification: The given equation is x2 0.4 = 0 put x = 0.2 in the given equation L.H.S. = (0.2)2 0.4 = 0.04 0.4 = 0.36 ≠ 0. 0.2 is not a solution of the given equation. 7. If b = 0, c < 0, is it true that the roots of
x2 + bx + c = 0 are numerically equal and opposite in sign? Justify.
Solution: Yes Justification: Consider the equation ax2 + bx + c = 0 Now, a = 1, b = 0, and c < 0 the equation becomes x2 c = 0 x2 = c x = c x = c the roots of the equation are numerically equal
and opposite in sign. 8. Does (x 1)2 + 2(x + 1) = 0 have real roots?
Justify your answer. Solution: No Justification: The given equation is (x 1)2 + 2(x + 1) = 0 x2 2x + 1 + 2x + 1 = 0 x2 + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 0, c = 1 b2 4ac = 0 4(1)(1) = 4 < 0 the roots of the given equation are not real. 9. Find whether the following equations have
real roots. If real roots exist, find them. i. 8x2 + 2x – 3 = 0 ii. 2x2 + 3x + 2 = 0 iii. 5x2 2x – 10 = 0
iv.
1
2 3x+
1
5x= 1, x
3,5
2 [CBSE 2011]
v. x2 + 5 5x 70 = 0 Solution: i. 8x2 + 2x – 3 = 0 Comparing with ax2 + bx + c = 0, we get a = 8, b = 2, c = –3 b2 – 4ac = (2)2 4(8)(3) = 4 + 96 = 100 > 0 Since, b2 4ac > 0 the roots are real and distinct The roots are given by
x = 2b b 4ac
2a
x = 2 100
2(8)
x = 2 10
16
x = 2 10
16
or x = 2 10
16
x = 8
16 or x = 12
16
x = 1
2 or x = 3
4
the roots of given the equation are 1
2,
3
4
.
ii. 2x2 + 3x + 2 = 0 Comparing with ax2 + bx + c = 0 , we get a = –2, b = 3, c = 2 b2 – 4ac = (3)2 – 4(2) (2) = 9 + 16 = 25 > 0 Since, b2 4ac > 0, the roots are real and distinct The roots are given by
x = 2b b 4ac
2a
x = 3 25
2( 2)
x = 3 5
4
x = 3 5
4
or x = 3 5
4
x = 2
4 or x = 8
4
x = 1
2
or x = 2
the roots of the given equation are 1
2
, 2.
iii. 5x2 2x – 10 = 0 Comparing with ax2 + bx + c = 0, we get a = 5, b = –2, c = –10 b2 4ac = (–2)2 4(5) (–10) = 4 + 200 = 204 > 0
42
Class X: Mathematics
42
Since, b2 – 4ac > 0, the roots are real and distinct The roots are given by
x = 2b b 4ac
2a
x = 2 204
2(5)
x = 2 2 51
10
x = 1 51
5
x = 1 51
5
or x = 1 51
5
the roots of given the equation are 1 51
5 5 ,
1 51
5 5 .
iv. 1
2 3x +
1
5x = 1
5 2 3
(2 3)( 5)
x x
x x
= 1
2
3 8
2 13 15
x
x x
=1
2x2 – 13x + 15 = 3x – 8 2x2 – 16x + 23 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = –16, c = 23 b2 – 4ac = (–16)2 – 4(2) (23) = 256 184 = 72 > 0 Since, b2 – 4ac > 0, the roots are real and distinct. The roots are given by
x = 2b b 4ac
2a
x =( 16) 72
2(2)
x = 16 6 2
4
x = 8 3 2
2
x = 8 3 2
2
or x =
8 3 2
2
x = 4 + 3 2
2or x = 4 –
3 2
2
the roots of the given equation are 4 +3 2
2,
4 – 3 2
2.
v. x2 + 5 5x 70 = 0 Comparing with ax2 + bx + c = 0, we get
a = 1, b = 5 5 , c = 70
b2 – 4ac = 2
5 5 4(1) (70)
= 125 + 280 = 405 > 0 Since, b2 4ac > 0, the roots are real and distinct The roots are given by
x = 2b b 4ac
2a
x =5 5 405
2(1)
x = 5 5 9 5
2
x = 5 5 9 5
2
or x =
5 5 9 5
2
x = 4 5
2or x =
14 5
2
x = 2 5 or x = 7 5
the roots of the given equation are 2 5 , 7 5. 1. Find the values of p for which the following
equation has equal roots. i. 9x2 + 8px + 16 = 0 ii. 4x2 + px + 3 = 0 [CBSE 2014]
iii. 2x2 + px + 9
2 = 0
vi. px(x 3) + 9 = 0 [CBSE 2014] v. (p + 4)x2 + ( p + 1) x + 1 = 0 vi. x2 + px + 16 = 0 2. Find value of k for which the following
equation has equal roots. i. x2 kx + (k 1) = 0 ii. kx2 2kx + 6 = 0 [CBSE 2012] iii. kx2 + 4x + 1 = 0 [CBSE 2012] iv. 9x2 3kx + k = 0 [CBSE 2014] v. (k 5)x2 + 2(k 5) x + 2 = 0 vi. x2 2x(1 + 3k) + 7(3+ 2k) = 0
[CBSE 2012] vii. (k 2)x2 + 2(2k 3)x + (5k 6) = 0
[CBSE 2015]
viii. kx (x 2 5 ) + 10 = 0 [CBSE 2013, 2014]
ix. x2 2(k + 1) x + k2 = 0 [CBSE 2013] x. x2 4kx + k = 0 [CBSE 2012]
Practice Problems based on Exercise 4.4
43
Chapter 04: Quadratic Equations
xi. k2x2 2 (2k 1) x + 4 = 0 xii. (k + 1) x2 2 (k 1) x + 1 = 0 xiii. (k 12) x2 + 2 (k 12) x + 2 = 0 xiv. (3k + 1) x2 + 2(k + 1)x + 1 = 0
[CBSE 2014] 3. If 5 is a root of the quadratic equation
2x2 + px 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots find the value of k.
[CBSE 2014, 2016] 4. If x = 2 is a root of the equation
3x2 + 7x + p = 0, find the value of k so that the roots of the equation x2 + k(4x + k 1) + p = 0 are equal.
[CBSE 2015] 5. If 1 is a root of the equation 3x2 + ax 2 = 0
and the quadratic equation a(x2 + 6x) b = 0 has equal roots, find the value of b.
[CBSE 2014] 6. If the root of the quadratic equation
(a b)x2 + (b c)x + (c a) = 0 a b are equal then prove that b + c = 2a.
[CBSE 2012, 2016] 7. If the roots of the equation
(a2 + b2)x2 2(ac + bd)x + (c2 + d2) = 0 are equal then prove that ad = bc.
[CBSE 2014] 8. Find the values of k for which the equation
(3k + 1)x2 + 2 (k + 1)x + 1 has equal roots. Also find the roots. [CBSE 2014]
9. Find the values of k for which the quadratic
equations (k + 4)x2 + (k + 1)x + 1 = 0 has equal roots. Also find roots.
[CBSE 2013, 2014] 10. If x = 4 is a root of the equation
x2 + 2x + 4p = 0, find the value of k for which the equation x2 + px(1 + 3k) + 7(3 + 2k) = 0 has equal roots. [CBSE 2015]
11. Find the nature of roots of the following
equations: i. x2 + 10x + 39 = 0
ii. 5x2 16 5 x + 4 = 0
iii. 4
5x2
2
7 x + 1 = 0
iv. 2x2 5x + 1 = 0 [CBSE 2011, 2012]
v. 8x2 22x 21 = 0
vi. 2
2 52 0
x x
vii. 4x2 20x + 25 = 0
12. The hypotenuse of right triangle is 3 5 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 15 cm. Is it possible to draw such a right triangle. If so, how long are the legs.
[CBSE 2012, 2014] 13. A pole has to be erected at a point on the
boundary of a circular park of diameter 15 m in such a way that difference from two diametrically opposite fixed gates P and Q on the boundary is 3 m. Is it possible to do so? If yes, at what distance from the two gates should the pole be erected?
1. i. 3, 3 ii. 4 3 , 4 3
iii. 6, 6 iv. 0, 4 v. 3, 5 vi. 8, 8 2. i. 2 ii. 6 iii. 4 iv. 0, 4
v. 7 vi. 2, 10
9
vii. 1, 3 viii. 2
ix. 1
2
x. 6
xi. 1
4 xii. 0, 3
xiii. 12, 14 xiv. 0, 1
3. 7
4
4. –1, 2
3
5. – 9
8. k = 0, 1. Roots are
(k 1) k k 1
3k 1
9. k = 5, 3. Roots are
(k 1) k 5 k 3
2 k 4
10. 2, 10
9
11. i. Not real ii. Real and distinct iii. Not real iv. Not real v. Real and distinct vi. Real and distinct vii. Real and equal 12. Yes. 3 cm, 6 cm 13. Yes. At 9 m and 12 m from the gates
Answers
44
Class X: Mathematics
44
1. The quadratic equation 2x2 – 5 x + 1 = 0 has [NCERT Exemplar] (A) two distinct real roots (B) two equal real roots (C) no real roots (D) more than 2 real roots 2. Which of the following equations has two
distinct real roots? [NCERT Exemplar]
(A) 2x2 – 3 2 x + 9
4 = 0
(B) x2 + x – 5 = 0
(C) x2 + 3x + 2 2 = 0 (D) 5x2 – 3x + 1 = 0 3. Which of the following equations has no real
roots? [NCERT Exemplar]
(A) x2 – 4x + 3 2 = 0
(B) x2 + 4x – 3 2 = 0
(C) x2 – 4x 3 2 = 0
(D) 3x2 + 4 3x + 4 = 0 4. Values of k for which the quadratic equation
2x2 – kx + k = 0 has equal roots is [NCERT Exemplar] (A) 0 only (B) 4 (C) 8 only (D) 0, 8 5. (x2 + 1)2 – x2 = 0 has [NCERT Exemplar] (A) four real roots (B) two real roots (C) no real roots (D) one real root 6. The value of k for which roots of quadratic
equation kx2 + 2x + 3 = 0 are equal is:
(A) 1
3 (B) 1
3
(C) 3 (D) – 3 7. The value of c for which the equation
ax2 + bx + c = 0 has equal roots is [CBSE 2012]
(A) 2b
a (B)
2b
4a
(C) 2a
b (D)
2a
4b
8. If the equation x2 – 4x + a = 0 has no real
roots, then: (A) a < 4 (B) a 4 (C) a < 2 (D) a > 4
9. Which of the following equations has the equal roots?
(A) x2 + 5x 500 = 0
(B) 2x2 – x + 1
8= 0
(C) 10x2 – 3x – 1 = 0 (D) 6x2 + 7x 10 = 0 10. Discriminant of the quadratic equation
5x2 + 3x – 7 = 0 is (A) 131 (B) 131 (C) 149 (D) –149
11. If the discriminant of the equation
6x2 – bx + 2 = 0 is 1, then the value of b is (A) 7 (B) 7
(C) 7 (D) 7 1. Check whether the given equation
(x – 3)2 + 9 = 2x2 is quadratic or not. Solution: (x – 3)2 + 9 = 2x2 x2 – 6x + 9 + 9 = 2x2 2x2 – x2 + 6x – 18 = 0 x2 + 6x – 18 = 0 which is of the from ax2 + bx + c = 0 and a 0 the given equation is quadratic. 2. If x = –2 is a solution of the quadratic
equation 2x2 + 3kx + 5 = 0, find the value of k.
Solution: The given equation is 2x2 + 3kx + 5 = 0 x = – 2 is root of the equation 2(–2)2 + 3k(–2) + 5 = 0 8 – 6k + 5 = 0 13 – 6k = 0 6k = 13
k 13
=6
the value of k is 13
6.
3. Find the roots of the quadratic equation 7x2 – 12x – 4 = 0. Solution: 7x2 – 12x – 4 = 0 7x2 – 14x + 2x – 4 = 0 7x(x – 2) + 2(x – 2) = 0 (x – 2)(7x + 2) = 0 x – 2 = 0 or 7x + 2 = 0
x = 2 or x = 2
7
the roots of the given equation are 2, 2
7
.
Multiple Choice Questions
One Mark Questions
45
Chapter 04: Quadratic Equations
4. Find the positive root of 24 + 48x = 8
Solution:
24 + 48x = 8
On squaring both sides, we get 4x2 + 48 = 64 4x2 = 64 48 4x2 = 16 x2 = 4 x = 2 the positive root of the given equation is 2. 5. If the roots of the quadratic equation
x2 + px + 25 = 0 are equal then find the value of p.
Solution: The given equation is x2 + px + 25 = 0 Since, the roots of the equation are equal, b2 – 4ac = 0 p2 – 4(1)(25) = 0 p2 – 100 = 0 p = 10 the value of p is 10 or 10. 6. Find the nature of roots for the quadratic
equation 5x2 – 7x + 3 = 0 Solution: 5x2 – 7x + 3 = 0 Comparing with ax2 + bx + c = 0, we get
a = 5, b = 7 , c = 3
b2 – 4ac = (– 7 )2 – 4(5)(3) = 7 – 60 = – 53 < 0 Since b2 – 4ac < 0, the roots of the given equation are not real. 1. For the quadratic equation given below, if
ps ≠ qr, then prove that the equation has no real roots.
x2(p2 + q2) + 2x (pr + qs) + (r2 + s2) = 0 Solution: The given equation is x2(p2 + q2) + 2x (pr + qs) + (r2 + s2) = 0 Comparing with ax2 + bx + c = 0, we get a = (p2 + q2), b = 2(pr + qs), c = (r2 + s2) b2 4ac = [2(pr + qs)]2 4(p2 + q2)(r2 + s2) = 4[(pr + qs)2
(p2r2 + p2s2 + q2r2 + q2s2)] = 4 (p2r2 + 2prqs + q2s2 – p2r2
p2s2 q2r2 q2s2) = 4 (2prqs p2s2 q2r2) = 4 (p2s2 2prqs + q2r2) b2 4ac = 4 (ps qr)2 ....(i)
But ps ≠ qr .... [given] ps qr ≠ 0 (ps qr)2 > 0 ...(ii) b2 4ac < 0 ...[From (i) and (ii)] the roots of the given equation are not real. 2. From a station, two trains start at the same
time. One train moves in west direction and other in North direction. First train moves 5 km/hr faster than the second train. If after 2 hours, distance between the two trains is 50 km, find the average speed of each train. [CBSE 2012]
Solution: Suppose that the two trains start from point O
and move in West and North respectively. After 2 hours, the two trains reach points A
and B respectively very such that AB = 50 km. Let the average speed of train moving in West
direction be x km/hr. Average speed of train moving in North
direction = (x + 5) km/hr. Distance covered by first train is 2 hrs = OA = 2x km Distance covered by second train in 2 hrs = OB = 2 (x + 5) km In right triangle OAB, by Pythagoras theorem, AB2 = OA2 + OB2 502 = (2x)2 + [2(x + 5)]2 2500 = 4x2 + 4 (x2 + 10x + 25) 2500 = 4 [x2 + x2 + 10x + 25] 625 = 2x2 + 10x + 25 2x2 + 10x 600 = 0 x2 + 5x 300 = 0 x2 + 20x 15x 300 = 0 x(x + 20) 15(x + 20) = 0 (x + 20) (x 15) = 0 x + 20 = 0 or x 15 = 0 x = 20 or x = 15 But speed cannot be negative. x = 15 km/hr Speed of other train = x + 5 = 15 + 5 = 20 km/hr the average speeds of the two trains are
15 km/hr and 20 km/hr.
HOTS Questions
W A x km/hr O
N
E
B
(x + 5) km/hr
S
50 km
46
Class X: Mathematics
46
3. If a, b, c, p, q and r are real numbers such that 2(ac + pr) = bq, then prove that atleast one of the equations ax2 + bx + c = 0 and px2 + qx + r = 0 has real roots.
Solution: Given equations are ax2 + bx + c = 0, and px2 + qx + r = 0 Let the discriminants of the above two
equations be D1 and D2 respectively. D1 = b2 4ac D2 = q2 4pr Now, D1 + D2 = b2 4ac + q2 4pr = b2 + q2 4(ac + pr) = b2 + q2 2 2 (ac + pr) = b2 + q2 2bq
....[ 2(ac + pr) = bq]
= (b q)2 ≥ 0
....[ (b q)2 ≥ for all real values of b and q]
D1 + D2 ≥ 0 D1 ≥ 0 or D2 ≥ 0 Atleast one equation has real roots. 4. The denominator of a fraction is one more
than thrice the numerator. If the sum of the
fraction and its reciprocal is 11
314
find the
fraction. Solution: Let the numerator of the fraction be x. Denominator = 3x + 1
Fraction = 3 1
x
x +
Reciprocal of the fraction = 3 1x +
x
According to the given condition,
3 1
x
x + +
3 1x +
x =
113
14
(3 1) 53
(3 1 14
x x +
x x +
2 2
)
2+ 9 + 6 1
3
2
2
x x x +
x x =
53
14
14(10x2 + 6x +1) = 53(3x2 + x) 140x2 + 84x + 14 = 159x2 + 53x 19x2 – 31x – 14 = 0 Comparing with ax2 + bx +c = 0, we get a = 19 , b = –31, c = –14 By quadratic formula,
x = 2b b 4ac
2a
x = 2( 31) ± ( 31) 4(19)( 14)
2(19)
x = 31 ± 961 + 1064
2(19)
x = 31 ± 2025
38
x = 31 ± 45
38
x = 76
38 or x =
14
38
x = 2 or x = 7
19
But x ≠ 7
19
x = 2
Fraction = 3 1
x
x+ = 2 2
6 + 1 7
the fraction is 2
7.
5. Determine p so that the equation
x2 + 7px + 4 = 0 has no real roots. Solution: The given equation is x2 + 7px + 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 7p and c = 4 Now b2 – 4ac = (7p)2 – 4(1)(4) = 49p2 – 16 Since the quadratic equation has no real
roots, b2 4ac < 0 49p2 – 16 < 0 49p2 < 16
p2 < 16
49
p2 16
49< 0
p2 2
4
7
< 0
4
p +7
4p
7
< 0
Either 4p
7 > 0 and 4
p +7
< 0
i.e. 4p >
7 and 4
p7
which is impossible.
Or
4
p7
< 0 and 4
p +7
> 0
i.e. 4p <
7 and 4
p7
4
7 < p < 4
7
47
Chapter 04: Quadratic Equations
Value Based Questions 1. The sum of the lives of CFL Lamp and
ordinary lamp is 16 years. Twice the square of the life of a CFL Lamp exceeds the square of the life of ordinary lamp by 164. Is this situation possible? If so, determine the life of each lamp. Which type of lamp is eco-friendly and one should prefer? Which value is depicted in the problem?
Solution: Let the life of CFL lamp be x years. Life of ordinary lamp = (16 – x) years According to the given condition, 2x2 = (16 x)2 + 164 2x2 (16 x)2 = 164 2x2 (256 32x + x2) = 164 2x2 256 + 32x x2 164 = 0 x2 + 32x 420 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 32, c = 420 b2 4ac = (32)2 4(1)( 420) = 1024 + 1680 = 2704 > 0 Since b2 4ac > 0, the roots are real and distinct the given situation is possible. The roots are given by
x = 2b b 4ac
2a
x = 32 2704
2(1)
x = 32 52
2
x = 32 52
2
or x =
32 52
2
x = 20
2or x =
84
2
x = 10 or x = 42 But life of lamp cannot be negative. x = 10 years Life of CFL lamp = 10 years Life of ordinary lamp = 16 x = 16 10 = 6 years One should prefer CFL lamp since it is eco-
friendly and has longer life The value depicted in the problem is i. Environmental awareness.
2. If the price of petrol in increased by ` 5 per liter, a person will have to buy 2 litre less petrol for ` 300. Find the original price and the increased price of petrol.
i. Why the price of petrol increasing day by day?
ii. Why should we conserve fuels? Solution:
Let the price of petrol be x per litre.
Quantity of petrol for ` 300 = 300
xlitres
New price of petrol = ` (x + 5)
Quantity of petrol for ` 300 = 300
5x+ litres
According to the given condition,
300
x –
300
5x+= 2
1 1
3005x x+
= 2
5
( + 5)
x + x
x x
=
2
300
2
5 1
5 150
x x
x2 + 5x = 750 x2 + 5x – 750 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 5, c = 750 By quadratic formula,
x = 2b b 4ac
2a
x = 25 5 4(1)( 750)
2(1)
x = 5 25 3000
2
x =
5 3025
2
x = 5 55
2
x = 5 55
2
or x =
5 55
2
x = 50
2 or x =
60
2
x = 25 or x = 30 But price of petrol cannot be negative. x = 25 Original price of petrol = ` 25 per litre Increased price of petrol = 25 + 5 = ` 30 per liter. i. The price of petrol increasing day by
day because of increased consumption of petrol.
ii. We should conserve fuels as it takes millions of years to form fossil fuels, and the reserves are getting depleted day by day.
48
Class X: Mathematics
48
Memory Map
Nature of roots of ax2 + bx + c = 0, a 0
Quadratic Equations Second degree polynomial
( General form: ax2 + bx + c = 0, a 0)
Method to solve Quadratic equation
ax2 + bx + c = 0, a 0
D = b2 – 4ac
Factorisation Completing the Square Quadratic formula
x = 2b ± b 4ac
2a
Steps
1. x2 + b
ax + c
a= 0
2. x2 + b
ax = c
a
3. x2 + b
22a
x
+ 2
b
2a
=2
b
2a
– c
a
4. 2 2
2
b b 4ac
2a 4ax +
5. 2± b 4acb
2a 2ax +
6. x = 2b ± b 4ac
2a
ax2 + bx + c = 0 (px + q) (rx + s) = 0
x = q
p , s
r
i. if D > 0, roots are real anddistinct
ii. if D = 0, roots are real andequal
iii. if D < 0, roots are not real
49
Chapter 04: Quadratic Equations
Sample Test Paper
Total Marks: 20
1. Find the roots of the equation x2 + 4x + 4 = 0 [1] 2. Check whether the equation (4x – 1) (x – 3) = (2x – 4) (2x – 3) is a quadratic or not [1] 3. Find the nature of the roots of the given quadratic equation, and find the roots if they exist
(x 5)(x 3) = 5 [2]
4. Find the roots of the quadratic equation 5x2 – 2 10 x + 2 = 0 [2]
5. Solve for x : 4 6 10
;5 7 3
x x
x xx 5, 7 [3]
6. If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2 – 2px + k = 0 has equal roots. Find the value of k. [3] 7. A two digited number is such that the product of its digits is 28. When 27 is added to this number,
the digits interchange their places. Find the number. [4] 8. A bus travel at a certain average speed for a distance of 75 km and then travels a distance of 90 km
at an average speed of 10 km/h more than the first speed. If it take 3 hours to complete the total journey, find its first speed. [4]