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124
An equation involving avariable of degree 2 isquadratic equation.
a
a
aa
m m
(m+2)
(m+2)
Consider a square of side ‘a’ units and its area 25 square units.
Area of the square = (side)2
25 = a2
or
a2 = 25 .............. (2)
In equation (2) the degree of the variable is two
What do you call such an equation?Such an equation is a quadratic equation.
a2 = 25
∴ a = ± 5
a = + 5 or a = –5
Consider a rectangle of sides ‘m’ and ‘(m + 2)’ units and its area is 8 sq units.
Area of a rectangle = (length) (breadth)
8 = (m) (m + 2)
8 = m2 + 2m
or
m2 + 2m = 8 .......................... (3)
Compare the equation (2) and (3)
In equation (2) a2 = 25, variable occurs only in second degree.
In equation (3) m2 + 2m = 8, variable occurs in second degree as well as in first degree.
Quadratic equation involving a variable only in second degree is a“Pure Quadratic Equation’’.
Example :
(1) x2 = 9 (2) 2a2 = 18
If the terms in the RHS are transposed to LHS then,
(1) x2 – 9 = 0 (2) 2a2 – 18 = 0
An equation that can be expressed in the form ax2 + c = 0, where a andc are real numbers and a ≠ 0 is a pure quadratic equation.
A Quadratic equation hasonly two roots.
125
Quadratic equation involving a variable in second degree as well as in firstdegree is an “Adfected Quadratic Equation”
Example :
(1) x2 + 3x = 10 (2) 3a2 – a = 2
If the terms in the RHS are transposed to LHS then,
(1) x2 + 3x – 10 = 0 (2) 3a2 – a – 2 = 0
ax2 + bx + c = 0 is the standard form of a quadratic equation where a, band c and variables and a ≠ 0.
1. Solving Pure Quadratic equationExample 1 : Solve the equation 3x2 – 27 = 0
Solution : 3x2 – 27 = 0
∴ 3x2 = 27
x� = 3
27
∴ x2 = 9
x = 9±x = +3 or x = –3
Example 2 : Solve the equation 4y2 – 9 = 0
Solution : 4y2 – 9 = 0∴ 4y2 = 9
∴ y2 = 4
9
y = 4
9±
y = 2
3±
y = 2
3+ or y = 2
3−
126
Example 3 : Solve the equation 99 = 4r2 – 1
Solution : 99 = 4r2 – 1
4r2 – 1 = 99
∴ 4r2 = 99 + 1
4r2 = 100
r2 = 4
100 = 25
∴ r = 25±
r = 5±r = +5 or r = –5
Example 4 : Solve the equation (m + 8)2 –5 = 31
Solution : (m + 8)2 –5 = 31
∴ (m + 8)2 = 31 + 5
(m + 8)2 = 36
∴ (m + 8)2 = 36
(m + 8) = 36± ∴ m = –8 ± 6
m = –8 + 6 or m = –8 – 6
m = –2 or m = – 14
Example 5 : If A = 2r� ; Solve for ‘r’.
Solution : A = 2rπ
2rπ = A
r2 = πA
r = π
± A
127
l2
Example 6 : If l2 = r2 + h2. Solve for h and find the value of ‘h’ if l = 15 and r = 9.
Solution : l2 = r2 + h2
or
∴ r2 + h2 = l2
∴ h2 = l2 – r2
∴ h = 22 rl −±
h = 22 915 −± (substituting l = 15, r = 9)
h = 81225 −±
h = 144± h = 12±∴ h = +12 or h = –12
Example 7 : If B = 4
a.3 2
Solve for ‘a’ and find the value of ‘a’ if B = 16 3 .
Solution : B = 4
a.3 2
∴ a2 = 3
B4
a = 3
B4± (Substituting B = 16 3 )
a = 3
3164
//×±
∴ a = 64± , a = ± 8
∴ a = + 8 or a = – 8
Exercise : 5.1
A. Classify the following equations into pure and adfected quadratic equation.
1) x2 + 2 = 6 2) a2 + 3 = 2a 3) p (p – 3) = 1
4) 2m2 = 72 5) k2 – k = 0 6) 7y = y
35
128
If mn = 0, then eitherm = 0 or n = 0
B. Solve the equations
1) 5x2 = 125 2) m2 – 1 = 143 3) 4a = a
81
4) 2
2x –
4
3 =
4
17 5) (2m – 5)2 = 81 6)
18
)4( 2−x =
9
2
C.
1) If A = 2 2rπ Solve for ‘r’ and find the value of ‘r’ if A = 77 and π = 7
22
2) If V = hr2π Solve for ‘r’ and find the value of ‘r’ if V = 176 and h = 14
3) If r2 = l2 + d2 Solve for ‘d’ and find the value of ‘d’ if r = 5 and l = 4.
4) If c2 = a2 + b2 Solve for ‘b’. If a = 8 and c = 17 and find the value of ‘b’.
5) If K = 1/2mv2 Solve for ‘v’ and find the value of ‘v’ if K = 100 and m = 2
6) If v2 = u2 + 2as. Solve for ‘v’. If u = 0, a = 2 and s = 100, find the valueof v.
2. Solving the adfected quadratic equation by factorization :Example 1 : Solve the quadratic equation a2 – 3a + 2 = 0
Solution : a2 – 3a + 2 = 0
i. Resolve the expression a2 – 2a – 1a + 2 = 0
ii. Factorize a(a – 2) –1 (a – 2) = 0
iii. Taking the common factor (a – 2) (a – 1) = 0
iv. Equate each factor to zero a – 2 = 0 or a – 1 = 0
v. The roots are a = 2 or a = 1
Example 2 : Solve the quadratic equation m2 – m = 6Solution : m2 – m = 6
∴ m2 – m – 6 = 0m2 – 3m + 2m – 6 = 0m(m – 3) +2 (m – 3) = 0(m – 3) (m + 2) = 0
Either (m – 3) = 0 or (m + 2) = 0m = +3 or m = –2
129
x2
1x 1x
Example 3 : Solve the quadratic equation 2x2 – 3x + 1 = 0Solution : 2x2 – 3x + 1 = 0
2x2 – 2x – 1x + 1 = 02x (x – 1) –1 (x – 1) = 0
∴ (x – 1) (2x – 1) = 0Either (x – 1) = 0 or (2x – 1) = 0
x = 1 or x = 2
1
Example 4 : Solve the quadratic equation 4k (3k – 1) = 5.
Solution : 4k (3k – 1) = 5
∴ 12k2 – 4k – 5 = 0
12k2 – 10k + 6k – 5 = 0
2k (6k – 5) + 1(6k – 5) = 0
∴ (6k – 5) (2k + 1) = 0
Either (6k – 5) = 0 or (2k + 1) = 0
k = 6
5or k =
2
1−
Exercise : 5.2A. Find the roots of the following equations
1) x(x – 3) = 0 2) a (a + 5) = 0 3) m2 – 4m = 0
4) 3k2 + 6k = 0 5) (y + 6) (y + 9) = 0 6) (b – 3) (b – 5) = 0
7) (2n + 1) (3n – 2) = 0 8) (5z – 2) (7z + 3) = 0
B. Solve the quadratic equations
1) x2 + 15x + 50 = 0 2) a2 – 5a + 6 = 0 3) y2 = y + 2
4) 6 – p2 = p 5) 30 = b2 – b 6) 2x2 + 5x – 12 = 0
7) 6y2 + y – 15 = 0 8) 6a2 + a = 5 9) 13m = 6(m2 + 1)
10) 0.2t2 – 0.04t = 0.03
Consider the equation x2 + 3x + 1 = 0It cannot be factorised by splitting the middle term.
How do you solve such an equation ?
It can be solved by using Formula.
130
3. Solving quadratic equation by formula method
General form of a quadratic equation ax2 + bx + c = 0
Divide by ‘a’ 0a
c
a
bx
a
ax2
=++
Transpose the constant term to R.H.S.a
c
a
bxx2 −=+
Add 2
a2
b
to both the sides22
2
a2
b
a
c
a2
b
a
bxx
+−=
++
2
22
a4
b
a
c
a2
bx +−=
+
2
22
a4
bac4
a2
bx
+−=
+
Simplify 2
22
a4
ac4b
a2
bx
−=
+
Taking square root 2
2
a4
ac4b
a2
bx
−±=+
a2
ac4b
a2
bx
2 −±=+
a2
ac4b
a2
bx
2 −±−=
∴ Roots area2
ac4bbx
2 −±−=
∴ a2
ac4bbx
2 −+−= ora2
ac4bbx
2 −−−=
Roots of the equation ax2 + bx + c = 0 are x = a2
ac4bb 2 −±−
Note : The roots of the equation ax2 + bx + c = 0 can also be found usingSridhara’s method.
x x
xx
xx
x
x
x
x
x x
x
x
x
131
Example 1 : Solve the equation x2 – 7x + 12 = 0
consider x2 – 7x + 12 = 0
This is in the form ax2 + bx + c = 0
the coefficients are a = 1, b = –7 & c = 12
The roots are given by x = a2
ac4bb 2 −±−
Substituting the values a = 1, b = –7 and c = 12
1x2
)12)(1(4)7()7(x
2 −−±−−=
2
48497x
−±=
Simplify x = 2
17 ±
x = 2
17 +or x =
2
17 −
x = 2
8or x =
2
6
Roots are x = 4 or x = 3
Example 2 : Solve the equation 2p2 – p = 15
Consider 2p2 – p = 15
2p2 – p – 15 = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 2, b = –1 and c = –15
The roots are given bya2
ac4bbx
2 −±−=
Substituting the values a = 2, b = –1 and c = –15
p = )2(2
)15)(2(4)1()1( 2 −−−±−−
x
x
x
x
x x
x
x x
132
x
p = 4
12011 +±+
p = 4
1211±
p = 4
111±
p = 4
111+or p =
4
111−
p = 4
12or p =
4
10−
p = 3 or p = 2
5−
Example 3 : Solve the equation 2k2 – 2k – 5 = 0Consider 2k2 – 2k – 5 = 0
This is in the form ax2 + bx + c = 0The coefficients are a = 2, b = –2 and c = –5
The roots are given by x = a2
ac4bb 2 −±−
Substituting the values a = 2, b = –2 and c = –5
k = )2(2
)5)(2(4)2()2( 2 −−−±−−
k = 4
4042 +± =
4
442 ±
k = 4
1122 ± =
( )4
1112 ±
∴ The roots are k = 2
111+ or k =
2
111−
133
Example 4 : Solve the equation m2 – 2m = 2Consider m2 – 2m = 2
m2 – 2m – 2 = 0This is in the form ax2 + bx + c = 0
Comparing the coefficients a = 1, b = –2 and c = –2
The roots are given by x = a2
ac4bb 2 −±−
m = )1(2
)2)(1(4)2()2( 2 −−−±−−
m = 2
842 +±+
m = 2
122 ±
m = ( )
2
312 ±
m = 31+ or m = 31−Exercise : 5.3Solve the following equations by using formula
1) a2 – 2a – 4 = 0 2) x2 – 8x + 1 = 0 3) m2 – 2m + 2 = 0
4) k2 – 6k = 1 5) 2y2 + 6y = 3 6) 8r2 = r + 2 7) p = 5 – 2p2
8) 2z2 + 7z + 4 = 0 9) 3b2 + 2b = 2 10) a2 = 4a + 6
4. Equations reducible to the form ax2 + bx + c = 0Example 1 : Solve the equation (x + 6) (x + 2) = x
Solution :(x + 6) (x + 2) = xx2 + 6x + 2x + 12 = xx2 + 8x + 12 – x = 0x2 + 7x + 12 = 0x2 + 4x + 3x + 12 = 0x(x + 4) + 3 (x + 4) = 0(x + 4) (x + 3) = 0
Either (x + 4) = 0 or (x + 3) = 0
x = – 4 or x = – 3
x
134
Example 2 : Solve the equation (a – 3)2 + (a + 1)2 = 16
Solution : (a – 3)2 + (a + 1)2 = 16
Using (a + b)2 = a2 + 2ab + b2
(a – b)2 = a� – 2ab + b2
[(a)2 – 2(a)(3) + 32] + [a2 + 2 (a) (1) + 12] = 16
a2 – 6a + 9 + a2 + 2a + 1 – 16 = 0
2a2 – 4a – 6 = 0
a2 – 2a – 3 = 0
a2 – 3a + 1a – 3 = 0
a(a – 3) + 1 (a – 3) = 0
(a – 3) (a + 1) = 0
Either (a – 3) = 0 or (a + 1) = 0
∴ a = 3 or a = – 1
Example 3 : Solve 5(p – 2)2 + 6 = 13 (p – 2)
Solution : 5(p – 2)2 + 6 = 13 (p – 2)
Let p – 2 = b
then 5b2 + 6 = 13b
5b2 – 13b + 6 = 0
5b2 – 10b – 3b + 6 = 0
5b (b – 2) –3 (b – 2) = 0
(b – 2) (5b – 3) = 0
Either (b – 2) = 0 or (5b – 3) = 0
b = 2 or b = 5
3
p – 2 = 2 or p – 2 = 5
3(∴ b = p – 2)
p = 2 + 2 or p = 5
3 +
1
2
p = 4 or p = 5
13
135
Example 4 : Solve the equation 1k
1k
5k2
2k3
−+=
++
Consider1k
1k
5k2
2k3
−+=
++
Cross multiplying (3k + 2) (k – 1) = (2k + 5) (k + 1)3k2 + 2k – 3k – 2 = 2k2 + 5k + 2k + 53k2 – 1k – 2 – 2k2 – 7k – 5 = 03k2 – 1k – 2 – 2k2 – 7k – 5 = 0
On simplification k2 – 8k – 7 = 0This is in form of ax2 + bx + c = 0
The co-efficients are a = 1, b = –8, c = –7
The roots of the equationa2
ac4)b(bx
2 −−+−=
∴ k = 1x2
)7)(1(4)8()8( 2 −−−±−−
∴ k = 2
28648 +±+
k = 2
928 ±
k = 2
2328 ± =
( )2
2342 ±
k = 234 ±
Example 5 : Solve the equation 1y2
3
4
y =−
Consider 1y2
3
4
y =−
Taking L.C.M. 1y4
6y2
=−
By cross multiplication y2 – 6 = 4yy2 – 4y – 6 = 0
x
136
this is in the form ax2 + bx + c = 0comparing coefficients a = 1, b = –4, c = –6
y = )1(2
)6)(1(4)4()4( 2 −−−±−−
the roots of the equation are = 2
24164 +±
y = 2
404 ±
y = 2
1024 ± =
( )2
1022 ±
y = 102 + or y = 102 −
Example 6 : Solve 1m2
4
3m
1
2m
4
+=
+−
+
1m2
4
)3m)(2m(
)2m(1)3m(4
+=
+++−+
1m2
4
6m3m2m
2m12m42 +
=+++
−−+
1m2
4
6m5m
10m32 +
=++
+
On Cross multiplying, 4m2 + 20m + 24 = 6m2 + 20m + 3m + 10
4m2 + 20m + 24 – 6m2 – 23m – 10 = 0
–2m2 – 3m + 14 = 0
This is in the Standard form 2m2 + 3m – 14 = 0
2m2 + 7m – 4m – 14 = 0
m(2m + 7) –2 (2m + 7) = 0
(2m + 7) (m – 2) = 0
Either (2m + 7) = 0 or (m – 2) = 0
m =2
7−or m = 2
137
Exercise : 5.4
A. Solve the following equations
1) (x + 4) (x – 4) = 6x 2) 2(a2 – 1) = a (1 – a)
3) 3(b – 5) (b – 7) = 4 (b + 3) 4) 8(s – 1) (s + 1) + 2 (s + 3) = 1
5) (n – 3)2 + n (n + 1)2 = 16 6) 11(m + 1) (m + 2) = 38 (m + 1) + 9m
B. Solve
1) 5
25
2
83
+−=
−−
x
x
x
x 2)
1a7
1a3
5a7
1a5
++=
++
3) 11m10
1m12
13m9
3m11
++=
++
4) 2y
)6y5)(4y(
3y
)6y5)(1y(
−+−=
−+−
5) xxx
2
1
2
2
1 =−
+−
6) 2b
8
b4
2
b5
3
+=
−+
− 7) 12
25
y
1y
1y
y =+++ 8)
1n
13n2
2n
2n
1n
1n
++=
−++
−+
9) 6m
6
)4m(2
5
2m
2
+=
++
+ 10) 22y
y5
3y4
)1y3(2 −+
=−−
5. To solve the problems based on Quadratic EquationExample 1 : If the square of a number is added to 3 times the number, the sum
is 28. Find the number.
Solution : Let the number be = x
Square of the number = x2
3 times the number = 3x
Square of a number + 3 times the number = 28
x2 + 3x = 28
x2 + 3x – 28 = 0
∴ x2 + 7x – 4x – 28 = 0
x(x + 7) –4 (x + 7) = 0
(x + 7) (x – 4) = 0
x + 7 = 0 or x – 4 = 0
x = –7 or x = 4
∴ The required number is 4 or –7
138
Example 2 : Sum of a number and its reciprocal is 55
1. Find the number.
Solution : Let the number be = y
Reciprocal of the number = y
1
(Number) + (its reciprocal) = 55
1
y + y
1 =
5
26
y
1y2 + =
5
26
5(y2 + 1) = 26y
5y2 + 5 = 26y
5y2 – 26y + 5 = 0
5y2 – 25y – 1y + 5 = 0
5y (y – 5) –1 (y – 5) = 0
(y – 5) (5y – 1) = 0
Either (y – 5) = 0 or (5y – 1) = 0
y = 5 or y = 5
1
∴ The required number is 5 or 5
1
Example 3 : The base of a triangle is 4 cms longer than its altitude. If the area ofthe traingle is 48 sq cms. Find the base and altitude.
Solution : Let the altitude = x cms.Base of the triangle = (x + 4) cms.
Area of the triangle = 2
1 (base) (height)
48 = 2
1 (x + 4)x
142
9) A dealer sells an article for Rs. 24 and gains as much percent as the cost priceof the article. Find the Cost price of the article.
10) Sowmya takes 6 days less than the number of days taken by Bhagya to completea piece of work. If both Sowmya and Bhagya together can complete the samework in 4 days. In how many days will Bhagya complete the work?
6. Nature of the roots of a quadratic equation.1) Consider the equation x2 – 2x + 1 = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 1, b = –2, c = 1
x = a2
ac4bb 2 −±−
x = 1x2
1x1.4)2()2( 2 −−+−−
x = 2
442 −±
x = 2
02 +
x = 2
02 +or x =
2
02 −
x = 1 or x = 1 → roots are equal
2) Consider the equation x2 – 2x – 3 = 0
This is in the form ax2 + bx + c = 0
the coefficients are a = 1, b = –2, c = –3
x = a2
ac4bb 2 −±−
x = 1x2
16)2( ±−−
x = 2
42 ±+
143
x = 2
42 +or x =
2
42 −
x = 2
6or x =
2
2−
x = 3 or x = –1 → roots are distinct
3) Consider the equation x2 – 2x + 3 = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 1, b = –2, c = 3
x = a2
ac4bb 2 −±−
x = 1x2
)3)(1(4)2()2( 22 −−±−−
x = 2
1242 −±
x = 2
82 −±
x = 2
222 −±
x = ( )
2
212 −± = 21 −±
x = 21 −+ or 21 −− → roots are imaginary
From the above examples it is clear that,1) Nature of the roots of quadratic equation depends upon the value of (b2 – 4ac)
2) The Expression (b2 – 4ac) is denoted by ∆ (delta) which determines the natureof the roots.
3) In the equation ax2 + bx + c = 0 the expression (b2 – 4ac) is called the discriminant.
144
Discriminant (b2 – 4ac) Nature of the roots
∆ = 0 Roots are real and equal
∆ > 0 (Positive) Roots are real and distinct
∆ < 0 (negative) Roots are imaginary
Example 1 : Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0.Consider the equation 2x2 – 5x – 1 = 0This is in form of ax2 + bx + c = 0The co-efficient are a = 2, b = –5, c = –1
∆ = b2 – 4ac∆ = (–5)2 –4(2) (–1)∆ = 25 + 8∆ = 33
∴ ∆ > 0Roots are real and distinct
Example 2 : Determine the nature of the roots of the equation 4x2 – 4x + 1 = 0Consider the equation 4x2 – 4x + 1 = 0This is in the form of ax2 + bx + c = 0The co-efficient are a = 4, b = –4, c = 1
∆ = b2 – 4ac∆ = (–4)2 –4 (4) (1)∆ = 16 – 16
∴ ∆ = 0 Roots are real and equal
Example 3 : For what values of ‘m’ roots of the equation x2 + mx + 4 = 0 are(i) equal (ii) distinctConsider the equation x2 + mx + 4 = 0This is in the form ax2 + bx + c = 0the co-efficients are a = 1, b = m, c = 4
∆ = b2 – 4ac∆ = m2 – 4(1) (4)∆ = m2 – 16
1) If roots are equal ∆ = 0 ∴ m2 – 16 = 0
m2 = 16
∴ m = 16 ∴ m = ± 4
145
2) If roots are distinct ∆ > 0 ∴ m2 – 16 > 0 ∴ m2 > 16
m2 > 16m > ± 4
Example 4 : Determine the value of ‘k’ for which the equation kx2 + 6x + 1 = 0 hasequal roots.
Consider the equation kx2 + 6x + 1 = 0
This is in the form ax2 + bx + c = 0
the co-efficients are a = k, b = 6, c = 1
∆ = b2 – 4ac
since the roots are equal, b2 – 4ac = 0 (∴ ∆ = 0)
(6)2 – 4(k)(1) = 0
36 – 4k = 0
4k = 36
k = 4
36 = 9
∴ k = 9
Example 5 : Find the value of ‘p’ for which the equation x2 – (p + 2) x + 4 = 0 hasequal roots.
Consider the equation x2 – (p + 2) x + 4 = 0
This is in the form ax2 + bx + c = 0
Coefficients are a = 1, b = –(p + 2), c = 4
since the roots are equal ∆ = 0b2 – 4ac = 0
[–(p + 2)]2 – 4(1)(4) = 0
(p + 2)2 – 16 = 0
p + 2 = ± 16
p + 2 = ± 4
p + 2 = + 4 or p + 2 = –4
∴ p = 4 – 2 or p = –4 – 2
∴ p = 2 or p = –6
146
If m and n are the roots of thequadratic equation
ax2 + bx + c = 0
Sum of the roots a
b−=
Product of roots a
c+=
Exercise : 5.6
A. Discuss the nature of roots of the following equations
1) y2 – 7y + 2 = 0 2) x2 – 2x + 3 = 0 3) 2n2 + 5n – 1 = 0
4) a2 + 4a + 4 = 0 5) x2 + 3x – 4 = 0 6) 3d2 – 2d + 1 = 0
B. For what positive values of ‘m’ roots of the following equations are
1) equal 2) distinct 3) imaginary
1) a2 – ma + 1 = 0 2) x2 – mx + 9 = 0
3) r2 – (m + 1) r + 4 = 0 4) mk2 – 3k + 1 = 0
C. Find the value of ‘p’ for which the quadratic equations have equal roots.
1) x2 – px + 9 = 0 2) 2a2 + 3a + p = 0 3) pk2 – 12k + 9 = 0
4) 2y2 – py + 1 = 0 5) (p + 1) n2 + 2(p + 3) n + (p + 8) = 0
6) (3p + 1)c2 + 2 (p + 1) c + p = 0
7. Relationship between the roots and co-efficient of the terms of the quadraticequation.
If ‘m’ and ‘n’ are the roots of the quadratic equation ax2 + bx + c = 0 then
m =a2
ac4bb 2 −+−, n =
a2
ac4bb 2 −−−
∴ m + n = a2
ac4bb 2 −+− +
a2
ac4bb 2 −−−
m + n = a2
ac4bbac4bb 22 −−−−+−
∴ m + n = a2
b2−
m + n = a
b-
mn =
−+−a2
ac4bb 2
−−−a2
ac4bb 2
mn = ( )
2
222
a4
ac4b)b( −−−
147
mn = ( )
2
22
a4
ac4bb −−
mn = 2
22
a4
ac4bb +−
∴ mn = 2a4
ac4 =
a
c∴ mn =
a
c
Example 1 : Find the sum and product of the roots of equation x2 + 2x + 1 = 0
x2 + 2x + 1 = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 1, b = 2, c = 1
Let the roots be m and n
i) Sum of the roots m + n = a
b− =
1
2−
∴ m + n = –2
ii) Product of the roots mn = a
c =
1
1
∴ mn = 1
Example 2 : Find the sum and product of the roots of equation 3x2 + 5 = 0
3x2 + 0x + 5 = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 3, b = 0, c = 5
Let the roots are p and q
i) Sum of the roots p + q = a
b− =
3
0
∴ p + q = 0
ii) Product of the roots pq = a
c =
3
5∴ pq =
3
5
148
Example 3 : Find the sum and product of the roots of equation 2m2 – 8m = 0
2m2 – 8m + 0 =0
The coefficients are a = 2, b = –8, c = 0
Let the roots be α and β
i) Sum of the rootsa
b−=β+α 2
)8(−−= = 4
ii) Product of the rootsa
c=αβ 2
0= = 0
Example 4 : Find the sum and product of the roots of equation x2 – (p+q)x + pq = 0
x2 – (p + q) x + pq = 0
The coefficients are a = 1, b = –(p + q), c = pq
i) Sum of the roots m + n = a
b−
m + n = ( )[ ]1
qp +−−
∴ m + n = (p + q)
ii) Product of the roots mn = a
c =
1
pq
∴ mn = pq
Exercise : 5.7Find the sum and product of the roots of the quadratic equation :1) x2 + 5x + 8 = 0 2) 3a2 – 10a – 5 = 0 3) 8m2 – m = 2
4) 6k2 – 3 = 0 5) pr2 = r – 5 6) x2 + (ab) x + (a + b) = 0
8. To form an equation for the given rootsLet ‘m’ and ‘n’ are the roots of the equation
∴ x = ‘m’ or x = ‘n’i.e., x – m = 0, x – n = 0
(x – m) (x – n) = 0 ∴ x2 – mx – nx + mn = 0
x2 – (m + n) x + mn = 0
149
If ‘m’ and ‘n’ are the roots then the Standard form of the equation isx2 – (Sum of the roots) x + Product of the roots = 0 x2 – (m + n) x + mn = 0
Example 1 : Form the quadratic equation whose roots are 2 and 3
Let ‘m’ and ‘n’ are the roots
∴ m = 2, n = 3
Sum of the roots = m + n = 2 + 3
∴ m + n = 5
Product of the roots = mn
= (2) (3)
∴ mn = 6
Standard form x2 – (m + n) x + mn = 0
x2 – (5)x + (6) = 0
∴ x2 – 5x + 6 = 0
Example 2 : Form the quadratic equation whose roots are 5
2 and
2
5
Let ‘m’ and ‘n’ are the roots
∴ m = 5
2 and n =
2
5
∴ Sum of the roots = m + n = 5
2 +
2
5 =
10
254 +
∴ m + n = 10
29
Product of the roots = mn = 2
5x
5
2∴ mn = 1
Standard form x2 – (m + n) x + mn = 0
∴ x2 –10
29x + 1 = 0
∴ 10x2 – 29x + 10 = 0
150
Example 3 : Form the quadratic equation whose roots are 3 + 2 5 and 3 – 2 5Let ‘m’ and ‘n’ are the roots
∴ m = 3 + 2 5 and n = 3 – 2 5Sum of the roots = m + n
= 3 + 2 5 + 3 – 2 5∴ m + n = 6
Product of the roots = mn
= (3 + 2 5 ) (3 – 2 5 )
= (3)2 –(2 5 )2
= 9 – 20∴ mn = – 11
x2 – (m + n) x + mn = 0 ∴ x2 – 6x – 11 = 0
Example 4 : If ‘m’ and ‘n’ are the roots of equation x2 – 3x + 1 = 0 find the value
of (i) m2n + mn2 (ii) n
1
m
1 +
Consider the equation x2 – 3x + 1 = 0This is in the form ax2 + bx + c = 0The coefficients are a = 1, b = –3, c = 1Let ‘m’ and ‘n’ are the roots
i) Sum of the roots m + n = a
b− =
1
)3(−− = 3
∴ m + n = 3
ii) Product of the roots mn = a
c
mn = 1
1∴ mn = 1
∴ (i) m2n + mn2 = mn (m + n)
= 1(3) = 3
(ii)m
1 +
n
1=
mn
mn + =
mn
nm + =
1
3
∴ m
1 +
n
1 = 3
151
Example 5 : If ‘m’ and ‘n’ are the roots of equation x2 – 3x + 4 = 0 form theequation whose roots are m2 and n2.
Consider the equation x2 – 3x + 4 = 0
The coefficients are a = 1, b = –3, c = 4
Let ‘m’ and ‘n’ are the roots
i) Sum of the roots = m + n = a
b− =
1
)3(−−
∴ m + n = 3
ii) Product of the roots = mn = a
c =
1
4
∴ mn = 4
If the roots are ‘m2’ and ‘n2’Sum of the roots m2 + n2 = (m + n)2 – 2mn
= (3)2 – 2(4)= 9 – 8
∴ m2 + n2 = 1
Product of the roots m2n2 = (mn)2
= 42
∴ m2n2 = 16
x2 – (m2 + n2) x + m2n2 = 0
∴ x2 – (1)x + (16) = 0
∴ x2 – x + 16 = 0
Example 6 : If one root of the equation x2 – 6x + q = 0 is twice the other, find thevalue of ‘q’
Consider the equation x2 – 6x + q = 0
This is in the form ax2 + bx + c = 0
The coefficients are a = 1, b = –6, c = q
Let the ‘m’ and ‘n’ are the roots
i) Sum of the roots m + n = a
b− =
1
)6(−−
∴ m + n = 6
152
ii) Product of the roots mn = a
c =
1
q
∴ mn = q
If one root is (m) then twice the root is (2m)
∴ m = m and n = 2m
m + n = 6
m + 2m = 6
3m = 6
∴ m =3
6∴ m = 2
We know that q = mn
q = m(2m)
q = 2m2
q = 2(2)2
q = 8∴ q = 8
Example 7 : Find the value of k so that the equation x2 – 2x + (k + 3) = 0 has oneroot equal to zero.
Consider the equation x2 – 2x + (k + 3) = 0
The coefficients are a = 1, b = –2, c = k + 3
Let ‘m’ and ‘n’ are the roots Product of the roots = mn
∴ mn = a
c
mn = 1
3k +
∴ mn = k + 3
Since ‘m’ and ‘n’ are the roots, and one root is zero then
m = m and n = 0 mn = k + 3
∴ m(0) = k + 3
∴ 0 = k + 3
∴ k = –3
153
Exercise : 5.8
A. Form the equation whose roots are
1) 3 and 5 2) 6 and –5 3) –2 and 2
34)
3
2 and
2
3
5) 2 + 3 and 2 – 3 6) –3 + 2 5 and –3 – 2 5
B.1) If ‘m’ and ‘n’ are the roots of the equation x2 – 6x + 2 = 0 find the value of
i) (m + n) mn ii) m
1 +
n
1
2) If ‘a’ and ‘b’ are the roots of the equation 3m2 = 6m + 5 find the value of
i) a
b
b
a + ii) (a + 2b) (2a + b)
3) If ‘p’ and ‘q’ are the roots of the equation 2a2 – 4a + 1 = 0 Find the value of
i) (p + q)2 + 4pq ii) p3 + q3
4) Form a quadratic equation whose roots are q
p and p
q
5) Find the value of ‘k’ so that the equation x2 + 4x + (k + 2) = 0 has one root equalto zero.
6) Find the value of ‘q’ so that the equation 2x2 – 3qx + 5q = 0 has one root whichis twice the other.
7) Find the value of ‘p’ so that the equation 4x2 – 8px + 9 = 0 has roots whosedifference is 4.
8) If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q
Graphical method of solving a Quadratic EquationLet us solve the equation x2 – 4 = 0 graphically,
x2 – 4 = 0
∴ x2 = 4
let y = x2 = 4
∴ y = x2
and y = 4
154
y = x2
x = 0 y = 02 y = 0
x = 1 y = 12 y = 1
x = 2 y = 22 y = 4
x = –1 y = (–1)2 y = 1
x = –2 y = (–2)2 y = 4
x 0 1 –1 2 –2 3y 0 2 2 8 8 6
(x, y) (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) ( 3 ,6)
Step 1: Form table ofcorresponding valuesof x and y
Satisfying the equationy = x2
Step 2: Choose the scale onx axis, 1 cm = 1 unity axis, 1 cm = 1 unit.
Step 3: Plot the points (0, 0);(1, 1); (–1, 1); (2, 4)and (–2, 4) on graphsheet.
Step 4: Join the points by asmooth curve.
Step 5: Draw the straight liney = 4 Parallel to x-axis
Step 6: From the intersectingpoints of the curve andthe line y = 4, drawperpendiculars to thex axis
Step 7: Roots of the equations are x = +2 or x = –2
The graph of a quadratic polynomial is a curve called ‘parabola’
Example 1 : Draw a graph of y = 2x2 and find the value of 3 , using the graph.
Step 1: Form the table ofcorresponding values ofx and y satisfying theequation y = 2x2
Step 2: Choose the scale on xaxis, 1 cm = 1 unit andy axis, 1 cm = 1 unit
Step 3: Plot the points (0, 0);(1, 2) (–1, 2); (2, 8) and(–2, 8) on graph sheet.
155
x 0 1 –1 2 –2
y 0 1 1 4 4
(x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)
x 0 1 –1 2 –2
y 2 1 3 0 4
(x, y) (0, 2) (1, 1) (–1, 3) (2, 0) (–2, 4)
Step 4: Join the points by asmooth curve
Step 5: Draw the straight liney = 6 Parallel to x-axis.
Step 6: From the intersectingpoints of the curve andthe line y = 6, drawperpendiculars to thex-axis.
Step 7: Value of 3 = ± 1.7
x = –1.7 or x = + 1.7
Example 2 : Draw a graph of y = x2 and y = 2-x and hence solve the equationx2 + x – 2 = 0
Step 1: Form the table ofcorresponding values ofx and y satisfying theequation y = x2
Step 2: Form the table ofcorresponding values ofx and y satisfying theequation y = 2 – x.
Step 3: Choose the scale on xaxis 1 cm = 1 unit andy axis, 1 cm = 1 unit.
Step 4: Plot the points (0, 0);(1, 1); (–1, 1); (2, 4)and (–2, 4) on the graphsheet.
Step 5: Join the points by asmooth curve.
Step 6: Plot the points (0, 2) ;(1, 1); (–1, 3); (2, 0)and (–2, 4) on graphsheet
156
x 0 1 –1 2 –2
y 2 1 1 4 4
(x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)
x 0 1 2 –1 –2
y 2 3 4 1 0
(x, y) (0, 2) (1, 3) (2, 4) (–1, 1) (–2, 0)
Step 7: Join the points to get a line.
Step 8: From the intersectingCurve and the line, drawperpendiculars to thex-axis
Step 9: Roots of the equation are ∴ x = 1 or x = –2
Example 3 : Solve the equation
Method I : x2 – x – 2 = 0
Split the equation
y = x2 and y = 2 + x
Step 1: Form the table ofcorresponding values xand y satisfying theequation y = x2
Step 2: Form the table ofcorresponding values xand y satisfying theequation y = 2 + x
Step 3: Choose the scale onx axis, 1 cm = 1 unity axis, 1 cm = 1 unit
Step 4: Plot the points (0, 0);(1, 1); (–1, 1); (2, 4)and (–2, 4) on the graphsheet.
Step 5: Join the points by asmooth curve
Step 6: Plot the points (0, 2);(1, 3) (2, 4); (–1, 1) and(–2, 0) on the graphsheet.
Step 7: Join the points to get astraight line
Step 8: From the intersectingpoints of Curve and theline, draw the perpendi-culars to the x-axis.
Step 9: Roots of the equation are x = –1 or x = 2
157
x 0 1 –1 2 –2
y –2 –2 0 0 4
(x, y) (0, –2) (1, –2) (–1, 0) (2, 0) (–2, 4)
Method II :
Step 1: Form the table ofcorresponding values ofx and y satisfyingequation y = x2 – x – 2.
Step 2: Choose the scale on xaxis 1 cm = 1 unit andy axis 1 cm = 1 unit.
Step 3: Plot the points (0, –2);(1 –2); (–1, 0); (2, 0)and (–2, 4) on the graphsheet.
Step 4: Join the points to forma smooth curve
Step 5: Mark the intersectingpoints of the curve andthe x – axis.
Step 6: Roots of the equations are x = –1 or x = 2
Exercise : 5.9
A. 1) Draw the graph of y = x2 and find the value of 7
2) Draw the graph of y = 2x2 and find the value of 3
3) Draw the graph of y = 2
1x2 and find the value of 10
B. 1) Draw the graph of y = x2 and y = 2x + 3 and hence solve the equationx2 – 2x – 3 = 0
2) Draw the graph of y = 2x2 and y = 3 – x and hence solve the equation2x2 + x – 3 = 0
3) Draw the graph of y = 2x2 and y = 3 + x and hence solve the equation2x2 – x – 3 = 0
C. Solve graphically
1) x2 + x – 12 = 0 2) x2 – 5x + 6 = 0 3) x2 + 2x – 8 = 04) x2 + x – 6 = 0 5) 2x2 – 3x – 5 = 0 6) 2x2 + 3x – 5 = 0