EQUATIONS AND INEQUATIONS

Linear Equations in Two Variables: The general form of a linear equation in two variables x and y is ax + by + c = 0, a 0, b 0. a, b, c being

real numbers.

To find a solution of a linear equation in two variables, we assign any value to one of the two variables and

then determine the value of the other variable from the given equation.

Illustration 1: x – y +7 = 0.

Solution: Taking x = 0, y = 7

x = 1, y = 8

- - - - - - - - - - -

- - - - - - - - - - - etc

There are infinite solutions.

Taking any two solutions, we can plot the corresponding points P and

Q. The line PQ is a graphical representation of the linear equation.

(7,O)

(O,7) P1

Q

(1,8)

Two properties of the graph are:

1. Every solution of the given equation determines a point on the line PQ.

2. Every point lying on the line PQ satisfies the given equation.

Illustration 2: Plot the graph of the following linear equation 3x + 2y = 9.

Solution: Get minimum two points and plot. X Y

0

1

2

3

4.5

3

1.5

0

0 1

2

3

4

1

2 3

y

x

x

x

Exercise 1: Plot the graph of the following linear equations

(i) y = x + 3 (ii) x =

(iii) y = –sin 3 (iv) (y – 2) = 3(x + 1) (v)

y 1 x 2=

5 1

Algebraic Solution of a system of Linear Equations in two variables: Equations of the form

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

form a system of linear equations in two variables.

Solution of a system of Linear Equations in two variables:

You are already familiar with the three methods

1. Method of elimination by substitution.

2. Elimination method by equating co-efficients.

3. Graphical Method.

There is one more method

4. Method of Cross Multiplication (Cramer’s Rule Method)

Let the given system of equations be a1x + b1y = c1

a2x + b2y = c2

Then 1 1

2 2

x

c b

c b

= 1 1

2 2

y

a c

a c

= 1 1

2 2

1

a b

a b

--------- (1)

Note: [This is known as Cramer’s method and1 1

2 2

a b

a bis called determinant,

a1

a2

b1

b2

D=

And its value is a1b2 – a2b1.].

From (1)

X =

1 1

2 2

1 1

2 2

c b

c b

a b

a b

= x 1 2 1 2

1 2 2 1

D c b b c

D a b a b

y =

1 1

2 2

1 1

2 2

a c

a c

a b

a b

= y 1 2 2 1

1 2 2 1

D a c a c

D a b a b

[Dx the determinant obtained by replacing the x – coefficients 1

2

a

a

with constant terms1

2

c

c

., Dy the

determinant obtained by replacing the coefficients 1

2

b

b

with constant terms1

2

c

c

].

Method of Solving: First write down the two equations with the constant terms on RHS.

Then find D, Dx and Dy and calculate x & y using the given formula.

Illustration 3: 2x + y = 35

3x + 4y = 65

Solution: 1 1

2 2

a b

a b

2

3

1

4

D =

= 8 – 3 = 5

Dx = 1 1

2 2

c b

c b =

35 1

65 4 = 140 – 65 = 75

Dy = 1 1

2 2

a c

a c =

2 35

3 65 = 130 – 105 = 25

Hence, x = xD 7515

D 5

Y =yD 25

5D 5

.

Exercise 2: Solve Using Cramer’s Method.

(i) 2x – y – 3 = 0 (ii) x y

+ = a + ba b

4x + y – 3 = 0 2 2

x y+ = 2

a b

(iii) (a – b)x + (a + b)y = 2a2 – 2b

2 (iv) a

2x + b

2y = c

2

(a + b)(x + y) = 4ab b2x + a

2y = d

2

(v) 2 3

+ = 13x y

5 4

= 2 [x 0, y 0]x y

Exercise 3: Solve

1 2+ = 3

x 1 y 1

4 5+ = 9

x 1 y 1

Exercise 4: Solve 3x 2y 1

+ 8 = 04 3 6

y x 13

+ 3 = 016 3 48

Conditions for consistency:

Consider the system of equations a1x + b1y = c1

a2x + b2y = c2

Case 1: The system of equations will have a unique solution and are said to be consistent if

1 1

1 2 2 1

2 2

a bD a b a b 0

a b

Such a system will graphically represent a pair of intersecting straight lines.

Eg 2x + 3y = 6

3x – y = 3

1 2 3 1

2

3

It can be seen that for this condition a1/a2 b1/b2

Case 2: The system of equations will have infinitely many solutions and are said to be dependent if

D = 0, Dx = 0 and Dy = 0. Or 1 1 1

2 2 2

a b c

a b c

This system of equations will graphically represent a pair of coincident lines as every solution of the

first equation is a solution of the second.

eg. –3x + 4y = –5

9

2 x – 6y =

15

2 Clearly 1 1 1

2 2 2

a b c

a b c .

Case 3: The system of equations will have no solution and are said to be inconsistent if

D = 0, and Dx or Dy 0.

i.e. 1 1 1

2 2 2

a b c

a b c .

Such a system of equations will represent a pair of parallel straight lines.

Eg 2x + 3y = 5 x y

+ = 63 2

.

Exercise 5: In each of the following system of equations determine whether the system has a unique solution,

no solution or infinitely many solutions. In case there is a unique solutions find it.

i) x – 3y = 3 ii) 3x – 5y = 20 iii) 2ax + 3by = a + 2b

3x – 9y = 2 21 35

x2 2

y =70 3ax + 2by = 2a + b

Exercise 6: Find the value of K for which the following system of equations has infinitely many solutions.

i) 4x + 5y = 3 ii) Kx 2y + 6 = 0

Kx + 15y = 9 4x 3y + 9 = 0

Exercise 7 : For what value of , the system of equations

x + 3y = 3

12x + y = will have no solutions?

Exercise 8: Prove that there is a value of c ( 0) for which the system

6x + 3y = c 3

12x + cy = c has infinitely many solutions. Find this value.

Exercise 9: Word Problems based on Linear equations in one and two variables.

i) Father and son have their ages in the ratio of 5: 2 . In 15 years their ages will be in the ratio 20 : 11. What

are their ages now.

ii) How soon afternoon will the hands of a clock be together again?

iii) If two liquids are mixed in the ratio of 3 : 2, a mixture is obtained weighing 1.04 g per cc while if they

bare mixed in the ratio 5 : 3, the resulting mixture weighs 1.05 g per c.c. Find the weight of a cc of each

of the original liquids.

iv) In a farm house there are rabbits and hens. They have between them 35 heads and 98 feet. How many

rabbits are there?

v) Find the positive integral solution of 13x + 15y = 189 for which the value of y is largest ?

vi) A coster manager sells half of his oranges plus half an orange, again he sells half of what remains plus

half an orange; and so again a third time. His stock is now exhausted, How many did he start with ?

Quadratic Equations Quadratic Polynomials: A polynomial of degree 2 is called a quadratic polynomial. Given form of quadratic

polynomial is ax2 + bx + c where a, b, c are real numbers such that a 0 and x is a variable.

Zeros or Roots of a Quadratic Polynomial: A real number is called a zero of a quadratic polynomial P(x)

= ax2 + bx + c if P() = 0.

Quadratic Equation: If P(x) is a quadratic polynomial, then P(x) = 0 is called a quadratic equation.

Roots of a Quadratic Equation: Let P(x) = 0 be a quadratic equation, then the zeros of the polynomial P(x)

are called the roots of the equation P(x) = 0, Thus x = is a root of P(x) = 0 if and only if P() = 0. A

quadratic polynomial may or may not have real zeros. A quadratic polynomial can have atmost two zeros or

two real roots.

Solutions of Quadratic Equations:

1) Factorization Method: 23x 10x 7 3 0

splitting the middle term

23x +3x +7x +7 3 = 0

3x(x 3) 7(x 3) 0

(x 3)( 3x 7) 0

Roots are 7

3 and3

Exercise 10: Solve:

i) x x + 1 34

+ =x + 1 x 15

x 0, x 1. ii)

1 2 6+ =

x 2 x 1 x.

iii) 2

4 3x + 5x 2 3 = 0

Exercise 11: Solve using square root :

(i)

2

2

1 13 x 16 x 26 = 0

xx (ii)

2

2

1 18 x 42 x 29 = 0

xx

(iii) 3x4 – 16x

3 + 26x

2 – 16x + 3 = 0 (iv) x

4 + x

3 – 4x

2 + x + 1 = 0

2. By Formula Method: Consider the quadratic equation ax

2 + bx + c = 0, a 0.

2 bx cx 0

a a [Dividing throughout by a]

2 b c

x xa a

2 2

2 b b c bx x

a 2a a 2a

[Add

2b

2a

on both sides to make a perfect square]

2 2

2

b b 4acx

2a 4a

2

b b 4acx

2a 2a

2b b 4ac

x2a 2a

2b b 4ac

2a

(b2 4ac) is called the discriminant of the quadratic equation and denoted by D. The nature of the roots

depends on the value of D.

Exercise 12: Solve for x

(i) 13 3x = x 3 (ii) x + 5 + x + 21 = 6x + 40

(iii) 3x2 – 4x + 4 – 4

23x 4x + 1 = 0 (iv)

x + 3 x 3 2x 3+ =

x + 2 x 2 x 1

Nature of the Roots of a Quadratic Equation:

The roots of the equation are real if b2 4ac 0. and complex if b

2 4ac < 0

Further if b2 4ac = 0, then the two roots are equal or coincident each = b/2a

If D is a perfect square, the roots are distinct and rational.

Summing Up:

D > 0, roots are real and distinct

D = 0, roots are real & coincident

D < 0, roots are complex.

Illustration 4: Find the values of the parameter a for which the quadratic equation (a + 1)x2 + 2(a + 1)x + a – 2 = 0.

(a) has two distinct roots,

(b) has no real roots,

(c) has two equal roots.

Solution: By the hypothesis this equation is quadratic, and therefore, a –1 and the discriminant of this

equation.

D = 4(a + 1)2 – 4(a + 1)(a – 2) = 4(a + 1)(a + 1 – a + 2) = 12(a + 1)

(a) For a > –1, then D > 0 then this equation has two distinct roots.

(b) For a < –1, then D < 0, then this equation has no real roots.

(c) This equation can not have two equal roots since, D = 0 only for a = –1, and this contradicts the

hypothesis.

Exercise 13: Find the nature of the roots of the following equations without actually solving it. (i) 2x

2 – 8x + 3 = 0

(ii) 4x2 – 20x + 25 = 0

(iii) x2 + 2(3a + 5)x + 2(9a

2 + 25) = 0

(iv) ax2 – 3bx – 4a = 0

(v) (x – a)(x – b) + (x – b)(x – c) + (x – c)(x – a) = 0

Exercise 14: Find the value of m for which the equation (1 + m)x

2 – 2(1 + 3m)x + (1 + 8m) = 0 has equal roots.

Exercise 15: If the roots of the equation (b – c)x

2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c.

Exercise 16: For what value of a, the equation (a

2 – a – 2)x

2 + (a

2 – 4)x + (a

2 – 3a + 2) = 0 will have more than two

solutions.

Exercise 17: Solve: 8

x + 5 = 7x + 5

.

Exercise 18: A length of 60 cm is divided into equal parts. What is the number of these parts if, when this

number is increased by unity, the length of each part is decreased by 1 mm?

Exercise 19: The distance from A to B by two different routes are 81 Km and 85 Km. A motor car taking the

longer road travels on the average 2 Km per hour faster than one taking the short route and does

the journey in 15 minutes less. Find the speed of each car.

Relation Between the Roots and Coefficients of the Quadratic Equation ax2 + bx + c = 0:

Roots of ax2 + bx + c = 0 are

2b b 4ac

2a 2a

Sum of Roots = + = b

a Product of Roots =

c

a.

To find equation whose roots are and

(x ) (x ) = 0

x2 – ( + )x + = 0

x2 (sum 0f roots)x + (product of roots) = 0

Illustration 5: If , are the roots of the equation (x – a)(x – b) = c, c 0. Then find the roots of the equation

(x – )(x – ) + c = 0

Solution: Since, , are the roots of

(x – a)(x – b) = c or x2 – (a + b)x + ab – c = 0

+ = a + b and = ab – c

Now (x – ) (x – ) + c = 0 x2 – ( + )x + + c = 0

x2 – (a + b)x + ab = 0 (x – a) (x – b) = 0 x = a, b

Hence, roots of (x – )(x – ) + c = 0 are a, b.

Exercise 20: If , are roots of x2

– x + 2 = 0 find the value of 3 +

3.

Exercise 21: If 3p2

= 5p + 2 and 3q2 = 5q + 2, ( p q). Find the equation whose roots are 3p – 2q and 3q – 2p.

Exercise 22: If one root of the equation px

2 – 14x + 8 = 0 is six times the another then find p.

Exercise 23: Prove that the roots of the quadratic equation ax23bx4a = 0 are real and distinct for all real a & b

Exercise 24: Prove that the roots of ax2 + bx + c = 0 will be real and distinct if and only if the roots of

(a + c) (ax2 + 2bx + c) = 2(ac b

2) (x

2 + 1) are imaginary.

Exercise 25: If one root of the equation x2+px+q=0 be the squares of the other, prove that p

3 q(3p 1) + q

2 = 0.

Exercise 26: Construct the equation whose roots are 2 + 5

2 and

2 5

2.

Exercise 27: If a and c are such that the quadratic equation ax2 5x + c = 0 has 10 as the sum of the roots and

also as the product of the roots, find a and c.

Exercise 28: If and are the roots of the quadratic equation ax2 + bx + c = 0, find the value of

3 +

3.

Condition for Common Roots Consider two quadratic equations:

ax2 + bx + c = 0 and ax

2 + bx + c = 0

(where a, a 0 and ab – ab 0)

Let be a common root then

a2 + b + c = 0 and a

2 + b + c = 0

Solving the above equations, we get

2 1

bc b c ca c a ab a b

From first two relations we get = bc b c

ca c a

and from last two relation we get = ca c a

ab a b

Eliminating , we get bc b c ca c a

ca c a ab a b

(bc – bc) (ab – ab) = (ca – ca)2

or

2a b b c c a

a b b c c a

(Remember)

This is the required condition for one root of two quadratic equations to be common.

If ab – ab = 0. then a b c

a b c

This is the required condition for both roots of two quadratic equations to be indetical.

Note: To find the common root between the two equations, make the same coefficient of x2 in both equations

and then subtract of the two equations.

Illustration 6: Find the value of k, so that the equations x2 – x – 12 = 0 and kx

2 + 10x + 3 = 0 may have one root

in common. Also find the common root.

Solution: Let be the common root of two equations.

then 2 – – 12 = 0 and k

2 + 10 + 3 = 0.

Solving the two equations:

2 1

117 12 3 10 k

......... (1)

(–12k – 3)2 = 117(1 + k)

9(4k + 1)2 = 117(10 + k)

(4k + 1)2 = 13(10 + k)

16k2 + 8k + 1 = 130 + 13k

16k2 – 5k – 129 = 0

16k2 – 48k + 43k – 129 = 0

or (k – 3) (16k + 43) = 0

k = 3 or k = –43/16

From (1) we have = 12k 3

10 +k

Hence, = –3 or 4

Exercise 29: For which value of k will the equations x2 – kx – 21 = 0 and x

2 – 3kx + 35 = 0 have one common root.

Exercise 30: Find the value of m so that the equation x2+ 10x + 21 = 0 and x

2 + 9x + m = 0 have a common root.

Find also the equation formed by the other roots.

Exercise 31: If the equations x2 + px + q = 0 and x

2 + rx + s = 0 have a common root, prove that it is equal to

ps rq q sas well as

q s r p. Find the remaining roots of the equation.

Linear Inequations: A statement of inequality between two expressions involving a single variable x with

highest power 1, is called a linear inequation.

e.g., (i) 2x + 5 < 3 (ii) 3x + 2 8 (iii) x

52 (iv) 4x 3 > 7

The general forms of linear inequations are:

(i)ax + b > c (ii)ax + b < c (iii)ax + b c (iv)ax + b c,

where a, b, c are real numbers and a 0.

Replacement Set or Domain of the Variable: The set from which the values of the variable x are

replaced in an inequation, is called the replacement set or the domain of the variable.

This replacement set is always given to us.

Solution Set: The set of all those values of x from the replacement set which satisfy the given inequation,

is called the solution of the inequation.

Solution set is always a subset of the replacement set.

Illustration 7: Write down the solution set of x < 5, when the replacement set is

(i) N (ii) W (iii) I.

Solution: (i) Solution Set = {x N : x < 5} = {1, 2, 3, 4}.

(ii) Solution Set = {x W : x < 5} = {0, 1, 2, 3, 4}.

(iii) Solution Set = {x I : x < 5} = {…, -2, -1, 0, 1, 2, 3, 4}.

Properties of Inequations:

i) Adding the same number of expression to each side of an inequation does not change the inequality.

ii) Subtracting the same number of expression from each side of an inequation does not change the

inequality.

iii) Multiplying (or dividing) each side of an inequation by the same positive number does not change

the inequality.

iv) Multiplying (or dividing) each side of an inequation by the same negative number reverses the

inequality.

e.g., –x < –2 x > 2 [Multiplying both sides by –1]

And, –2x < 8 x > –4 [Dividing both sides by –2]

Remarks: (i) a < b b > a

(ii) a > b b < a

Thus, x > 3 3 < x and x < 5 5 > x.

Some Special Sets of Numbers Shown on Number Line

We would like to have a glimpse on how we represent sets of numbers on number line.

Some sets of numbers and their graphs are given below.

Graphs of Subsets of N, W and I:

i) Graph of {x: -1 < x < 4, x N} = {1, 2, 3}.

2 1 2 3

The darkened circles indicate the natural numbers contained in the set.

ii) Graph of {x: -1 < x < 4, x W} = {0, 1, 2, 3}.

2 1 0 1 2 3

The darkened circles indicate the whole numbers contained in the set.

iii) Graph of {x : x > 2, x N} = {3, 4, 5, 6, ………}

0 1 2 3 4 5 6 1

The darkened circles indicate the natural numbers contained in the set and three dark dots above

the right part of the line show that natural numbers are continued indefinitely.

iv) Graph of {x : x < -1, x } = {-2, -3, -4, -5, …….}.

5 4 3 2 1 0 1

The darkened circles show the integers contained in the set and three dark dots above the left part

of the line show the indefinite continuity of negative integers.

Graphs of Sets of All Real Numbers Between Two Given Numbers:

We show the end points of the set by two circles, namely a hollow circle for the number not contained in the

set and the darkened circle for the number contained in the set. And, the line segment between these circles

is darkened.

v) Graph of {x : 1 x 3, x R}

2 1 0 1 2 3 4 5

Note that the end points 1 and 3 are both contained in the set.

vi) Graph of {x : 2 x < 3, x R}

3 2 1 0 1 2 3 4

Note that 2 is there in the given set while 3 is not contained in it.

vii) Graph of {x : 1 < x < 5, x R}.

1 0 1 2 3 4 5 6

Here 1 as well as 5 is not contained in the given set.

viii) Graph of {x : 0 < x 4}.

2 1 0 1 2 3 4 5 6

Here 0 is not contained in the set while 4 is contained in it.

Graphs of {x : x > a, x R} and {x : x a, x R} :

Such a set has one end point. The ray to the right of this point is darkened. If the end point is contained in

the set, then it is shown by a darkened circle, other wise by a hollow circle.

ix) Graph of {x : x 2, x R}.

1 0 1 2 3 4

Note that the end point 2 is contained in the given set.

x) Graph of {x : x > 3, x R}.

1 0 1 2 3 4 5

Here the end point 3 is not contained in the set.

Graphs of {x : x < a, x R} and {x : x a, x R} :

Such a set has one end point. The ray to the left of this end point is darkened. The end point is shown by a

darkened or a hollow circle, according as it is contained or not contained in the set.

xi) Graph of {x : x 3, x R}.

1 0 1 2 4 5 2 3

Here the end point 3 is contained in the set.

xii) Graph of {x : x < 1, x R}.

3 2 1 0 1 2 3

Here the end point 1 is not contained in the set.

Illustration 8: State giving reasons whether the following statements are true or false:

(i) If a > b, then a c > b c;

(ii) If a < b, then ac < bc;

(iii) If a > b, then a b

>c c

;

(iv) If a c < b d, then a + d < b + c;

where a, b, c, d are real numbers and c 0.

Solution: We have:

(i) a > b a c > b c [Adding c on both sides]

If a > b, then a c > b c is a true statement.

(ii) a < b ac < bc is true only when c > 0.

If a < b, then ac < bc is not true for all nonzero value of c.

(iii) a > b a b

>c c

is true only when c > 0.

If a > b, then a b

>c c

is not true for all non zero values of c.

(iv) a c < b d (a c) + (d + c) < (b d) + (d + c)

[adding (d + c) on both sides]

a + d < b + c.

If a c < b d, then a + d < b + c is a true statement.

Illustration 9: Find the solution set of 3(x 2) < 1, where x {1, 2, 3, 4, 5, 6}.

Solution: We have

3(x 2) < 1 3x 6 < 1

3x < 7 [Adding 6 on both sides]

x < 7

3 [Dividing both sides by 3]

x < 1

23

.

Solution set = 1

x : x < 2 ,x A ,3

where A = {1, 2, 3, 4, 5, 6} = {1, 2}.

Illustration 10: Solve the inequation 5

12 +1 x 5 + 3x6

, x R. Represent the solution set on a number line.

Solution: We have 5 11

12 1 x 5 3x 12 x 5 3x6 6

72 + 11x 30 + 18x [Multiplying both sides by 6]

11x 18x 42 [Adding 72 on both sides]

7x 42 [Adding 18x to both sides]

x 6 [Dividing both sides by 7]

Solution set = {x : x 6, x R}.

This set can be represented on the number line, as shown below.

1 0 1 2 3 4 5 6 7

Illustration 11: Solve the inequation 3 3 2x < 9, x R. Represent the solution set on a number line.

Solution: 3 3 2x < 9 3 3 2x and 3 2x < 9,

Now, 3 3 2x 6 2x [Adding 3 to both sides]

2x 6

x 3 …(i) [Dividing both sides by 2]

Again 3 2x < 9 2x < 6 [Adding 3 on both sides]

x > 3 [Dividing both sides by 2]

3 < x …(ii)

From (i) and (ii), we get { 3 < x 3, x R}.

This set can be represented on the number line as shown below.

4 3 2 1 0 1 2 3 4

Illustration 12: Solve the equation 2x – 3 < x + 2 3x + 5, x R. Represent the solution set on the number line.

Solution: 2x – 3 < x + 2 3x + 5 2x – 3 < x + 2 and x + 2 3x + 5.

Now, 2x – 3 < x + 2 x – 3 < 2 [Adding – x to both sides]

x < 5 {Adding 3 to both sides]

Again, x + 2 3x + 5 3x + 5 x + 2 …(i)

2x + 5 2 [Adding – x to both sides]

2x – 3 [Adding – 5 to both sides]

x –3/2 [Dividing both sides by 2]

–3/2 x ……….(ii)

From (i) and (ii), we get –3/2 x < 5.

Solution set =3

x : x 5,x R2

.

This set can be represented on the number line, as shown below:

3 –2 –1 0 2 3 4 –3/2 1 5

Illustration 13: Write down the range of real values of x for which the inequations x > 3 and –2 x < 5 are both

true.

Solution: Clearly we have to find the common values of the solution sets of the given inequations.

The graphs of these inequations are given below.

2 -1 0 1 2 3 4

2 -1 0 1 2 3 4 5

The common portion of the two graphs is the set {x : 3 < x < 5, x R}.

Hence, the required set is {x : 3 < x < 5, x R}.

Illustration 14: Solve and graph the solution set of 3x – 4 > 11 or 5 – 2x 7.

Solution: We have:

3x – 4 > 11 3x > 15 [Adding 4 to both sides]

x > 5 [Dividing both sides by 3]

Let A = {x : x > 5, x R}.

Again, 5 – 2x 7 –2x 2 [Adding – 5 to both sides]

x – 1 [Dividing both sides by – 2]

Let B = {x : x – 1, x R}

Required Solution Set = A B

= {x R : x > 5} {x R : x – 1}.

The graph of the set is given below:

1 0 1 2 3 4 5

Illustration 15: The following diagram represents two inequations A and B on number lines.

1 0 1 2 3 4 5 6 A =

1 0 1 2 3 4 5 6 7 8 B =

(i) Write down A, B and B in set builder form.

(ii) Represent A B and A B on two different number lines.

Solution: (i) Clearly, we have:

A = {x R : – 1 x < 5}, B = {x R : 2 x < 7}.

B = { x R : x < 2 } { x R : x 7 }.

(ii) From the graphs, it is clear that

A B = {x R : 2 x < 5} And A B = {x R : – 1 x < 2}.

The graphs of A B and A B may be drawn, as shown below:

1 0 1 2 3 4 5 6 A B =

1 0 1 2 3 4 5 6 7 8 A B` =

Exercise 32: Solve each of the inequations given below and represent its solution set on a number line:

(i) 2x – 7 < 4, x {1, 2, 3, 4, 5, 6, 7}

(ii) 4 – 3x 3x – 14, x N

(iii) 30 – 2(3x – 4) < 24, x W

(iv) –3 < 2x – 1 < x + 4, x I

(v) 2 + 4x < 2x – 5 < 3x, x I

Exercise 33: Find the smallest value of x, which satisfies the inequation 7 5x

2x + > + 3,2 3

x .

Exercise 34: Solve each of the inequations given below and graph the solution set on the number line:

(i) –4 < 3x + 2 11, x R

(ii) 1 15 – 7x > 2x – 27, x N

Exercise 35: Find the range of values of x, which satisfy 1 x 1 1

1 <3 2 3 6 , Graph the values of x on the

real line.

Exercise 36: Find the range of values of x, which satisfy 2 1 1

2 x + < 33 3 3 , x R.

Graph the values of x on the number line.

Exercise 37: Given : P = {x : 5 < 2x – 1 11, x R}

And Q = {x : – 1 3 + 4x < 23, x I },

Where R = {real numbers} and I = {ingers}.

Represent P and Q on he number line. Write down the elements of P Q.

Exercise 38: Solve each of the following inequations and graph the solution set on the number line:

(i) 5x – 11 7x – 5 < 9 (ii) 2x – 1 7 x

x +3

> 2

Exercise 39: The given diagram represents two sets A and B on the number line:

–2 –3 –1 0 –4 3 2 4 5 1

–2 –3 –1 0 3 2 4 5 1

(i) Write down A and B in set builder form.

(ii) Write down A B, A B, A B, A – B and B – A and represent them on separate

number lines.

Exercise 40: Given : A = { x : 5x – 4 6, x R } and B = { x : 5 – x > 1, x R }

Represent A and B on the real line. Find (i) A B (ii) A B

Quadratic Inequalities in one Variable:

We shall be concerned with the equations ax2 + bx + c = 0 (a 0) and the related inequalities ax

2+bx+c>0;

ax2 + bx + c < 0 where a > 0 and a, b, c are real. Only real values of x will be considered.

Recall that the solutions of ax2 + bx + c = 0 are real and different (unequal) if b

2 – 4ac > 0

Then we can write ax2 + bx + c = a(x – r1)(x – r2) where r1 , r2 are real and r1 r2.

Suppose r1 < r2.

Let us assume x is moving from left to right along the real number line and consider how the sign of

ax2+bx+c varies in the process.

At the extreme left (where x < r1 < r2) then (x – r1) is negative and (x – r2) is negative.

Hence, in this region a(x - r1)(x – r2) is positive. When x reaches r1, a(x – r1)(x – r2) becomes zero. Between

r1 and r2, (x – r1) is positive and (x – r2) is negative. So in this region a (x – r1) (x – r2) is negative. At r2 a (x –

r1) (x – r2) is zero. To the right of r2, (x – r1) and (x – r2) are both positive, so that a (x – r1) (x – r2) is positive.

These results can be summerised in the following figure.

Here the solutions of ax2 + bx + c = 0 are real and unequal since a > 0 and r1 and r2 real, r1 r2.

0 w –

r2

+

r1

0 + –

+, -, +, 0 relate to the sign of the expression ax2 + bx + c.

If ax2 + bx + c = 0 (a > 0) has real roots and unequal real solutions r1 and r2. Then when

x < r1, ax2 + bx + c > 0

x = r1, ax2 + bx + c = 0

r1 < x < r2, ax2 + bx + c < 0

x = r2, ax2 + bx + c = 0

x > r2, ax2 + bx + c > 0

Similarly we can show the following results:

(A) If ax2 + bx + c = 0 (a > 0) has equal real solutions r1 = r2 = r (say) then when

x < r ax2 + bx + c > 0

x = r ax2 + bx + c = 0

x > r ax2 + bx + c > 0

(B) If ax2 + bx + c = 0 (a > 0) has non real solutions then for all real x, ax

2 + bx + c > 0

Solution of Quadratic Inequations: The following illustration illustrates the method of solving a quadratic

inequation.

Illustration 16: Solve x2 – 6x + 8 > 0.

Solution: x2 – 6x + 8 > 0

i.e., (x – 2) (x – 4) > 0

(x – 2) < 0 and (x – 4) < 0

OR (x – 2) > 0 and (x – 4) > 0

x < 2 and x < 4

OR x > 2 and x > 4

x < 2 OR x > 4

The set of solutions of x2 6x + 8 > 0 is {x : x < 2 or x > 4}.

Illustration 17: Solve x2 6x + 5 < 0.

Solution: x2 6x + 5 < 0

(x 1)(x 5) < 0

(x 1) > 0 and (x 5) < 0

OR

(x 1) < 0 and (x 5) > 0 (why ?)

x > 1 and x < 5

OR

x < 1 and x > 5

(1 < x < 5)

The set of solutions

of x2 6x + 5< 0 is {x :1 < x < 5}

x > 1 x < 5

5 0 1

Exercise 41: Solve the following inequations

i) x2 x 2 < 0 ii) x

2 4x 21 > 0

iii) x2 11x + 10 > 0 iv) x

2 + 9x 22 < 0

The Graphs of Quadratic Functions:

To draw the graph of quadratic functions y = ax2 + bx + c, we give to x various values, both positive and

negative and find the corresponding values of y. The sets of corresponding values of x and y which give rise

to points are plotted on graph paper. The points are all joined by a smooth curve. This curve represents the

function y = ax2 + bx + c.

Illustration 18: Sketch the graphs of the functions y = x2 and y = x

2 on one and the same graph.

y = x2

x 0 2 2 3 3 5 5

y 0 4 4 9 9 25 25

y = x2

x 0 2 2 3 3 5 5

y 0 4 4 9 9 25 25

Solution: Scale: Along x axis : 1 division = 1 unit

Along y axis : 1 division = 5 units.

-5 -4 -3 -2 -1 5 4 3 2 1

5

10

15

20

25

-5

-15

-20

-25

-10

-30

30

0 x`

1. Both the curves pass through origin.

2. y = x2 and y = x

2 are symmetrical about y axis.

3. For y = x2, 0 is the minimum value of y.

4. For y = x2, 0 is the maximum value of y.

5. (0, 0) is the turning point for both the graphs.

The curves are parabolas.

Graphical Solutions of Equations:

Illustration 19: Draw the graph of the function y = x2 4x, y R and hence solve the equation x

2 4x = 0.

Solution:

y = x2 4x

x 2 1 0 1 2 3 4 5

y = x2 4x 12 5 0 3 4 3 0 5

Scale: Along x axis : 1 division = 1 unit

Along y axis : 1 division = 2 units

-5 -4 -3 -2 -1 1 2 3 4 5 6 x x

-2

-4

-6

Y`

2

4

6

8

10

12

14

16

Y

The solutions of the equation x2 4x = 0 are got by putting y = 0 in equation y = x

2 4x.

The solutions of the equation are the points of intersection of the curve y = x2 4x with v=0 i.e., the axis

The solutions are x = 0 and x = 4

Illustration 20: Sketch the graph of the function y = x2 + x 12 and hence solve the equation x

2 + x 12 = 0.

Solution: The tabulated values of the function of x2 + x 12 are given below:

x 5 4 1 0 1 3 4

y = x2 + x 12 8 0 12 12 10 0 8

Scale: Along x axis : 1 division = 1 unit

Along y axis : 1 division = 2 units

-5 -4 -3 -2 -1 0 5 4 3 2 1

-6

-4

-8

-12

-10

-2

2

4

6

8

x x`

y

y`

The curve is a parabola.

The solution of the equation x2 + x 12 = 0 are got by putting y = 0 in equation y = x

2 + x 12.

The solutions are the points of intersection of the curve y = x2 + x 12 with y = 0 i.e., the x axis. The

solutions are x = 4 and x = 3.

Exercise 42: Find graphically the roots of the equation 7 + 4x – x2 = 0and maximum values of the function

7 + 4x – x2.

Exercise 43: Draw the graph of y = x2 + 2x – 8 and from the graph, determine the minimum value of

x2 + 2x – 8 and roots of x

2 + 2x – 8 = 0.

Answers to exercise

2. (i) x = 1, y = 1 (ii) x = a2, y = b

2 (iii)

2 22ab a bx

b

,

3 3 2 2a b ab a by

b(a b)

(iv) 2 2 2 2

4 4

a c b dx

a b

,

2 2 2 2

4 4

a d b cy

a b

(v) x = ½, y = 1/3 3. x = 2, y = 2

4. x = 10, y = 1

5. (i) No Solution (ii) Infinite Solutions (iii) 4a b

x5a

,

a 4by

5b

6. (i) K = 12 (ii) K = 8/3

7. = 6 8. c = 6

9. (i) 45 yr & 18 yr (ii) 5

511

min (iii) 1.20 g ; 0.80 g (iv) 14 Rabbits, 21 Hens

(v) x = 3, y = 10 (vi) 7 Oranges

10.(i) x = 3/2 , 5/2 (ii) x = 3, x = 4/3 (iii) 3 2

,4 3

11. (i) x = 1, 3, 1/3 (ii) x = 4, –¼, 2, –½ (iii) 0, 1, 3, 1/3 (iv) x = 1, 1,3 5

2

12. (i) x = 4 (ii) x = 4 (iii) x = 4/3, x = 0 (iv) x = 0, x = 4

13. (i) real, distinct (ii) equal (iii) no real roots (iv) real, distinct

(v) real, distinct

14. m = 0, m = 3 16. a = 2 17. x = –6, 3 18. 24

19. 18km, 20km 20. –6, 21. 3x2 – 5x – 100 = 0 22. p = 3, 0

26. x2 – 2x – ¼ = 0 27. a = ½ , c = 5 28.

3

3

3abc b

a

29. k = 4

30. m = 18, x2 + 13x + 42 = 0; m = 14, x

2 + 5x + 6 = 0 31.

q(r p) s(r p),

q s q s

32.

1 –1 0 2 3 4 5 6

1 –1 0 2 3

1 0 2 3 4 5 6

1 0 2 3 4

–4 –5 –3 –2 –1 0 1

33. x = – 1

34.

1 –2 –1 0 2 3

3 0 1 2 4 5

35. { x : 2 x < 3, x R}

2 1 3 4 0

36. { x : – 3 x < 3, x R}

0 –2 –1 2 3 1 –3

37. P = { x : 3 < x 6, x R }

3 0 1 2 4 5 6

Q = { – 1, 0, 1, 2, 3, 4 }

2 –1 0 1 3 4

+

P Q = { 4 }

38.(i) { x : – 3 x < 2, x R }

0 –3 –2 –1 1 2

(ii) { x : x 5/2, x R }

3 0 1 2

39.(i) A = { x : – 2 < x 3, x R } and B + { x : 0 x < 5, x R }

(ii) A B = { x : – 2 < x < 5, x R}, A B = { x : 0 x 3, x R}

(iii) A B = { x : 3 < x < 5, x R }, A – B = { x : – 2 < x < 0, x R }

and B – A = { x : 3 < x < 5, x R }

(A B)

2 –2 –1 0 1 3 4 5

3

(A B) 0 1 2

4

(A` B) (B - A) 0 1 2 3 5

0 1 2 -3 -2 -1

(A – B)

40.

0 1 2 3 4 -5 -4 -3 -2 -1 A

0 1 2 3 4 B

41. (i) –1 < x < 2, (ii) x < –3, x > 7 (iii) x < 1, x > 10 (iv) –11 < x < 2

ASSIGNMENTS

SUBJECTIVE

LEVEL – I

1. The coefficient of x in the quadratic equation ax2 + bx + c = 0 was wrongly taken as 17 in place of 13

and its roots were found to be 2 and 15. Find the actual roots of the equation.

2. For the given system of equations kx – y – 2 = 0, 6x – 2y – 3 = 0 determine the value of K for which

the given system of equations has (a) a unique solution (b) no solution. Is there a value of K for

which the system has infinitely many solutions?

3. Solve: bx + cy = a + b

1 1 1 1 2a

ax cya b a b b a b a a b

4. If one root of the quadratic equation ax2 + bx + c = 0 is double the other, prove that 2b

2 = 9ac.

5. A stream flows from A to B a distance of 30 km at 2 km/hr and a man can now up and down in 8

hours. Find the rate of man in still water.

6. Solve the equation 3x2 11x 12 = 0 graphically, taking values from 5 to 6.

7. In a ABC, A = x , B = 3x and C = y. If 3y 5x = 30, prove that the triangle is right angled.

8. Solve the equation x 1 2x 5 3 .

9. Solve the inequation given below and graph the solution set on the number line

1 2x 5

2 12 3 6

, x .

10. Solve x2 + 3x 18 < 0.

LEVEL – II

1. Solve the equation

3x – 16y – 3 = 0

2x + 5y – 49 = 0

2. Solve the equation

32x

+ 3x+1

= 5 – 3x

3. A cyclist after riding a certain distance stopped for half an hour to repair his machine, after which he

completes the whole journey of 30 Km at half speed in 5 hours. If the breakdown had occurred 10

Km further on, he would have done the journey in 4 hours. Find where the breakdown occurred and

his original speed.

4. A polygon of n sides has n(n 3)

2

diagonals. How many sides has a polygon with 54 diagonals?

5. Determine m so that the two equations mx + a = bx + m

ax + m = mx + b are consistent.

OBJECTIVE

LEVEL – I 1. If the roots of the equation ax

2 + bx + c = 0 are equal, then each root is equal to

(A) c/a (B) –b/a (C) –b/2a (D) –c/a 2. The sum of the roots of a quadratic equation is –7/2 and product is 5/2. The equation is (A) 2x

2 + 7x – 5 = 0 (B) 2x

2 – 7x + 5 = 0

(C) x2 +

7x

2 + 5 = 0 (D) 2x

2 + 7x + 5 = 0

3. x2 – 5x + 6 < 0

(A) x > 3 (B) x < 2 (C) x lies between 2 and 3 (D) x does not lies between 2 and 3

4. The graph of y = x

2 – x – 12 meets the x–axis at

(A) (4, 0), (3, 0) (B) (4, 0), (–3, 0) (C) (4, –3) (D) (0, 4), (0, –3)

5. The solution set of 3x 13 = x + 1 is

(A) {4} (B) {4, –3} (C) {2} (D) {1, –1} 6. If the given system of the equations kx + 3y – (k – 3) = 0 12x + ky – k = 0 has infinitely many solutions if k = (A) 3 (B) 2 (C) 6 (D) None of these 7. The system of equation 5x + 2y = 16

15

x 3y 242

(A) has a unique solution (B) has no solution (C) has infinitely many solutions (D) are inconsistent

8. If 10 – 5x < 5(x + 6), x N the smallest value of x is (A) x = –1 (B) x = 0 (C) x = 1 (D) None of these 9. If – 3 is one root of the equation x

2 + kx – 6 = 0, then the value of k is

(A) 1 (B) 3 (C) 2 (D) – 1 10. The roots of x

2/3 + x

1/3 – 2 = 0 are

(A) 1, 8 (B) 2, 4 (C) –8, 1 (D) –2, 4

LEVEL – II

1. If the product of the roots of the equation x2 + bx +

2 + 1 = 0 is – 2 then equals

(A) –1 (B) 1 (C) 2 (D) – 2

2. If the product of the roots of the equation x2 – 2 2 kx + 2k

2 – 1 = 0 is 31, then the roots of the

equation are real for k equal to (A) –1 (B) 2 (C) 3 (D) 4

3. If and are the roots of the equation ax2 + bx + c = 0, then the value of

2 +

2 + is

(A) c(a – b)/a2 (B) 0

(C) –bc/a2 (D) none of these

4. If (m

2 – 3)x

2 + 3mx + 3m + 1 = 0 has roots which are reciprocals of each other, then the value of m

equals to (A) 4 (B) –3 (C) 2 (D) none of these

5. If 4 x 9 then

(A) (x – 4)(x – 4) 0 (B) (x – 4)(x – 9) 0

(C) (x – 4)(x – 9) 0 (D) (x – 4)(x – 9) > 0

6. If , are the roots of the equation x2 + x + = 0 then the values of and are

(A) = 1, = –1 (B) = 1, = –2

(C) = 2, = 1 (D) = 2, = –2 7. If the roots of the equation x

2 – 2ax + a

2 + a – 3 = 0 are real and less than 3, then

(A) a < 2 (B) 2 a 3

(C) 3 < a 4 (D) a > 4

8. The sum of the roots of 1 1 1

x a x b c

is zero. The products of roots is

(A) 0 (B) 1

2(a + b)

(C) –1

2(a

2 + b

2) (D) 2(a

2 + b

2)

9. If p and q are the roots of the equation x

2 + px + q = 0, then

(A) p = 1, q = –2 (B) p = 0, q = 1 (C) p = –2, q = 0 (D) p = –2, q = 1

10. The largest negative integer which satisfies 2x 1

0(x 2)(x 3)

is

(A) –4 (B) –3 (C) –1 (D) –2

ANSWERS

SUBJECTIVE

LEVEL – I 1. –3, –10

2. (a) k 3 (b) k = 3 (c) no value of k gives infinitely many solutions 3. x = a/b y = b/c 5. 8 km/hr

8. x = 3 9. 15

2,4

10. – 6 < x < 3.

LEVEL – II 1. x = 17, y = 3 2. x = 0 3. 15 Km from starting point, 10 Km/Hr 4. n = 12

5. m = b+a/2 if b a : If b = a, both the equations reduce to the same equation. This equation is consistent for all values of m.

OBJECTIVE

LEVEL – I 1. C 2. D

3. C 4. B

5. A 6. C

7. C 8. C

9. A 10. C

LEVEL – II

1. A 2. D

3. A 4. A

5. B 6. B

7. A 8. C

9. A 10. D