18681858 math o level solution of quadratic equation

Chapter No. 1

SOLUTION OF QUADRATIC EQUATION

EXERCISE 1AQ.1 Solve the following quadratic equations by factorisation:(a) x2 5x = 0Solution:

x2 5x = 0Take x as common

x(x 5) = 0Either

x = 0or

x 5 = 0x = 5

So, x = 0 or x = 5(b) 4x2 = 7x Solution:

4x2 = 7x 4x2 7x = 0

Take x as commonx(4x 1) = 0

Eitherx = 0

or4x 7 = 0

x = 47

2 Emporium Career Maths Series

x = 143

So, x = 0, or x = 143

(c) 6t2 = t(6 4)Solution:

6t2 = t(t 4)6t2 t(t 4)

6t2 t2 + 4t = 05t2 + 4t = 0

t(5t + 4) = 0Either

t = 0or

5t + 4 = 0

t = 54

So, t = 0, or to = 54

(d) 5y2 = y(y + 3)Solution:

5y2 = y(y + 3)5y2 = y2 + 3y5y2 y2 3y = 0

y(4y 3) = 0Either

y = 0or

4y 3 = 0

y = 43

So, y = 0 or y = 43

Emporium Career Maths Series 3

(e) a2 + 9a = 0Solution:

a2 + 9a = 0a(a+9) = 0

Eithera = 0

ora + 9 = 0

a = 9So, a = 0 or 9

(f) 3h2 = h(5 2h)Solution:

3h2 = h(5 2h)3h2 (5 2h) = 0 3h2 + 2h2 5h = 0

5h2 5h = 05h(h 1) = 0

Eitherh = 0

orh 1 = 0

h = 1So, h = 0 or 1

(g) x2 2x + 1 = 0Solution:

x2 2x + 1 = 0 x2 x x + 1 = 0

x(x 1) 1 (x 1) = 0(x 1)2 = 0

Eitherx 1 = 0

x = 1or


x 1 = 0x = 1

So, x = 1, 1(h) 7a + a2 18 = 0 Solution:

7a + a2 18 = 0Re-arranging

a2 + 7a 18 = 0a2 + 9a 2a 18 = 0a(a+9) 2(a+9) = 0

(a+9) (a2) = 0Either

a + 9 = 0a = 9

ora 2 = 0

a = 2So, a = 9 or a = 2

(i) 2x2 + 5z 3 = 0Solution:

2z2 + 5z 3 = 02z2 + 6z z 3 = 0

2z(z+3) 1(z3) = 0(z+3) (2z1) = 0

Eitherz + 3 = 0

z = 3or

2z 1 = 0

z = 21

So, z = 3 or z = 21


(j) c2 + 2c = 35 Solution:

c2 + 2c = 35c2 + 2c 35 = 0

c2 + 7c 5c 35 = 0c(c+7) 5(c7) = 0

(c+7) (c5) = 0Either

c + 7 = 0c = 7

orc 5 = 0

c = 5So, c = 7 or c = 5

(k) 8p 16 p2 = 0 8p 16 p2 = 0

Re-arrangingp2 8pr + 16 = 0

p2 4p 4p + 16 = 0p(p4) (p4) = 0

Eitherp 4 = 0

p = 4or

p 4 = 0p = 4

So, p = 4 or 4(l) 4 3b b2 = 0Solution:

4 3b b2 = 0b3 + 3b 4 = 0

b2 + 4b b 4 = 0b(b+4) 1(b+4) = 0


(b+4) (b1) = 0Either

b + 4 = 0b = 4

orb 1 = 0

b = 1So, b = 4 or b = 1

(m) 12 a a2 = 012 a a2 = 0a2 + a 12 = 0

a2 + 4a 3a 12 = 0a(a+4) 3(a+4) = 0

(a+4)(a3) = 0Either

a + 4 = 0a = 4

ora 3 = 0

a = 3So, a = 4, or a = 3

(n) 10t2 t = 2Solution:

10t2 t = 210t2 t 2 = 0

10t2 5t + 4t 2 = 05t(2t1) +2(2t1) = 0

(2t1)(5t+2) = 0Either

2t 1 = 0

t = 21

or


5t + 2 = 0

t = 52

So, t = 21

or = 52

(o) y2 22y + 96 = 0Solution:

y2 22y + 96 = 0y2 16y 6y + 96 = 0y(y16) 6(y16) = 0

(y16) (y6) = 0Either

y 16 = 0y = 16

ory 6 = 0

y = 6So, y = 16 or y = 6

(p) 12a2 16a 35 = 012a2 16a 35 = 0

12a2 30a + 14a 35 = 06a(2a5) +7(2a5) = 0

(2a 5) (6a + 7) = bEither

2a 5 = 0

a = 221

or6a + 7 = 0

6a = 7

a = 67


a = 161

So, a = 221

or a = 161

(q) 15x2 + 4x 35 = 0Solution:

15x2 + 4x 35 = 015x2 + 25x 21x 35 = 0

5x(3x+5) 7(3x+5) = 0(3x+5) (5x7) = 0

Either3x + 5 = 0

x = 35

x = 132

or5x 7 = 0

x = 57

x = 152

So, x = 132

or a = 152

(r) 15x2 + 4x 35 = 0Solution:

28x2 85x + 63 = 028x2 36x 49x + 63 = 0

4x(7x9) 7(7x9) = 0(7x9) (4x7) = 0

Either7x 9 = 0, 7x = 9


x = 79

x = 172

or4x 7 = 0, 4x = 7

x = 47

x = 143

So, x = 172

or x = 143

(s) 56x2 159x + 108 = 0Solution:

56x2 159x + 108 = 056x2 96x 63x + 108 = 08x(7x12) 9(7x12) = 0

(7x12) (8x9) = 0Either

7x 12 = 0 7x = 12

x = 7

12

x = 175

or8x 9 = 0, 8x = 9

x = 89

x = 181

So, x = 175

or x = 181

(t) 39x2 = 131x 44 = 0


Solution:39x2 + 131x 44 = 0

39x2 + 143x 12x 44 = 013(3x11) 4(3x11) = 0

Either3x 11 = 0 3x = 11

x = 332

or13x 4 = 0 13x = 4

x = 134

So, x = 321

or x = 134

(u) 76x 96x2 15 = 0Solution:

76x 96x2 15 = 096x2 76x + 15 = 0

96x2 40x 36x + 15 = 08x(12x5) 3(12x5) = 0

(12x5) (8x3) = 0Either

12x 5 = 0

x = 125

or8x 3 = 0

x = 83

So, x = 125

or 83

Q. Form a quadratic equation in x with the giver roots for each of the following:


(a) 2, 3Solution:

Given roots arex = 2 and x = 3

x 2 = 0 x 3 = 0 the required equation is

(x 2) (x 3) = 0x2 2x 3x + 6 = 0

x2 5x + 6 = 0(b) 3, 4Solution:


x3 = 0 x+4 = 0 the required equation is

(x3) (x+4) = 0x2 3x + 4x 12 = 0

x2 + x 12 = 0(c) 5, 0Solution:


x + 5 = 0 x 6 = 0 the required equation is

(x + 5) (x 6) = 0x2 x 30 = 0

(d) 5, 21

Solution:

x = 5 and x = 21

x 5 = 0 2x 1 = 0 the required equation is


(x 5) (2x 1) = 02x2 x 10x + 5 = 0

2x2 11x + 5 = 0

(e) 32 , 5

4

Solution:Given roots are

x = 32

and x = 54

3x 2 = 0 5x + 4 = 0 the required equation is

(3x 2) (5x + 4) = 015x2 + 12x 10x 8 = 0

15x2 + 2x 8 = 0

(f) 87 , 8

5

Solution:Given roots are

x = 87

and x = 65

8x = 7 6x = 58x + 7 = 0 6x 5 = 0

So our required equation is(8x + 7) (6x 5) = 0

48x2 40x + 42x 35 = 048x2 + 2x 35 = 0

(g) 2 21 , 1 4

3

Solution:Given factors

x = 221

and x = 1 43


x = 25

, 2xn = 5 x = 47

x,n = 7

2x + 5 = 0 4x 7 = 0So our required equation is

(2x + 5) (4x + 7) = 08x2 + 14x + 20x + 35 = 0

8x2 + 34x + 35 = 0

(h) 1 21 , 3

2

Solution:Given factors

x = 121

and x = 32

, 3n = 2

x = 23

, 2n = 3 3x = 2

2x + 3 = 0 3x 2 = 0So our required equation is

(2x + 3) (3x 2) = 06x2 4x + 9x 6 = 0

6x2 + 5x 6 = 0

(i) 74 , 7

4

Solution:Given the factors

x = 74

and x = 74

7x + 4 = 0 7x + 4 = 0So our required equation is,

(7x + 4) (7x + 4) = 049x2 + 28x + 28x + 16 = 0

49x2 + 56x + 16 = 0


EXERCISE No. 1BQ.1. Solve the following equations, giving your answers correct

to 2 decimal places where necessary.(a) (x + 1)2 = 9 Solution:

(x + 1) = 9Taking the square root of each side.Either,

x + 1 = 3x = 2

orx + 1 = 3

x = 4(b) (2x + 1)2 = 16Solution:

(2x + 1) = 16Taking the square root of each side.Either,

2x + 1 = 4

x = 23

x = 1.5or

2x + 1 = 4

x = 25

x = 2.5(c) (3x + 2)2 = 49Solution:

(3x + 2) = 49Taking the square root of each side.Either.

3x + 2 7


x = 35

x = 1.67or

3x + 2 = 7x = 3

(d)2

432

+x = 4529

Solution:Taking the square root of each side.Either,

2x + 43

= 57

2x = 57

43

2x = 201528

x = 4013

x = 0.33or

2x + 43

= 57

2x = 57

43

2x = 201528

x = 4043

x = 1.08(e) (5x 4)2 = 81Solution:

(5x 4) = 81


Taking square root of each side.Either.

5x 4 = 9

x = 5

13

x = 2.60or

5x 4 = 9

x = 55

x = 1

(f) (3x +7)2 = 4925

Solution:Taking square root of each side.Either,

3x + 7 = 75

3x = 75

7

3x = 7495

x = 2144

x = 2.10or

3x + 7 = 75

3x = 75

7

3x = 7

595


x = 2154

x = 2.57(g) (x-4)2 = 17

(x 4)2 = 7 Taking square rot of each side.Either,

x 5 = 4.12x = 8.12

orx 4 = 4.12

x = 0.12(h) (x + 3)2 = 11Solution:

(x + 3y) = 11Taking square root of each side.Either,

x + 3 = 3.32x = 0.32

orx + 3 = 3.32

x = 6.32(i) (2x 3)2 = 23Solution:

(2x 3) = 23Taking square root of each side.Either,

2x 3 = 4.80x = 3.90

or2x 3 = 4.80

x = 0.90(j) (3x +2)2 = 43Solution:


(3x + 2) = 43Taking square root of each side.Either,

3x + 2 = 6.65x = 1.32

or3x + 2 = 6.56

x = 2.85(k) (5x 7)2 = 74Solution:

(5x 7) = 74Taking square root of each side.Either,

5x 7 = 8.6x = 3.12

or5x 7 = 8.6

x = 0.32(l) (3 + 7x)2 = 65Solution:

(3 + 7x)2 = 65Taking square root of each side.Either,

3 + 7x = 8.067x = 8.06 37x = 5.06x = 0.72

or3 + 7x = 8.06

7x = 8.06 37x = 11.06x = 1.38

Q.2 What number must be added to each of the following expressions to obtain a perfect square?


(a) x2 + 7x Solution:

Given expression is x2 + 7x The coefficient of x is 7.So square of half of will be added both side.

x2 + 7x + 2

27

= 2

27x

+

2

27

must be added

(b) x2 3x Solution:

Given expression is x2 3x The coefficient of x is 3So square of half of 3 will be added both side.

x2 3x + 2

23

= 2

23x

2

23

must be added

(c) x2 + 27 x

Solution:

Given expression is x2 + 27 x

The coefficient of x is 27

So square of half of 27 must be added.

x2 + 27 x +

2

27

=

+27x

2

27

=

1649

must be added.


(d) x2 1.8x Solution:

Given expression is x2 1.8x

The coefficient of x is 1.8 = 59

So square of half of = 59 will be added both side.

x2 59

x +

10

9=

2

109x

10081

must be added.

(e) a2 + 2.4aSolution:

Given the expression is a2 + 2.4n

The coefficient of a is 2.4 = 5

12

So square of half of 5

12 will be added both side.

x2 + 5

12a +

2

56

= 2

56a

+

2536

must be added.

(f) c2 + 432 c

Solution:

Given the expression is c2 + 432

The coefficient of c is 3

14.

So square of half of 314 will be added both side.

c2 +3

14c +

2

37

= 2

37c

+


2

37

=

949

must be added.

(g) y2 + 54 y

Solution:

Given the express in is y2 = 54

y

The coefficient of y is 54

.

So squared half of 54 will be added both side.

y2 + 5y

y + 2

52

= 2

52y

+

254

must be added.

(h) v2 321 v

Solution:

Given expression is v2 321

v

The coefficient of v is 27

p2 27

v + 2

47

= 22

47v

2

47

or

1649

must be added.

(i) b2 10kbSolution:

Given the expression is b2 10kb

The coefficient of b is 1-k. Half of this is 2

k10 = 5k.

b2 10kb + (5k)2 = (b 5k)2


25k2 must be added.(j) g2 5kgSolution:

Given expression is g2 5kg

The coefficient of g is 5k. Half of this is 2

k5.

g2 5kg + 2

2k5

= 2

2k5y

4k25 2 must be added.

(k) h2 + 3mhSolution:

Given expression is h2 +3mh

The coefficient of h is 3 m. Half of this is 2m3

.

h2 + 3mh + 2

2m3

= 2

2m3h

+

4m9 2 must be added.

(l) k2 131 k

Solution:

Given expression is k2 131

k.

The coefficient of k is 34

. Half of this is 32

k2 34

+ 2

32

= 2

32k

94

must be added.

(m) d2 + 10xdSolution:


Given expression is a2 + 10xdThe coefficient of d is 10x. Half of this is 5x.

D2 + 10xd + (5x)2 = (d + 5x)2

(5x)2 = 25x2 must be added.(n) h2 5xkSolution:

Given expression is k2 5xkThis is a quadratic expression in k.

The coefficient of k is 5x. Half of this is 2

x5.

k2 5xk +

2

x5=

2

2x5x

4x25 2 must be added.

(o) m2 5n2mSolution:

Given expression is m2 5n2mThis is a quadratic expression in m.

The coefficient of m is 5n2. Half of this is 25

n2.

m2 5n2m + 22

2n5

= 22

2n5m

4n25 4 must be added.

EXERCISE NO.1CQ. Solve the following equations by factorisation where

possible or by completing the square. If the answers involve decimal places, give them correct to 2 decimal places. If an equation has no real roots, indicate that this is so.

(a) x2 + 2x + 3 = 0Solution:

x2 + 2xz + 3 = 0


x2 + 2x = 3x2 + 2x + (1)2 = 3 + (1)2

(x + 1)2 = 2Either,

x + 1 = + 2or

x + 1 = 2Hence, roots of x2 + 2x + 3 = 0 are complex.

(b) 5x2 4x 2 = 0Solution:

5x2 4x 2 = 0

x2 54

x 52

= 0

x2 54

x = 52

x2 54

x + 2

52

=

52

+ 2

52

2

52x

= 1514

Either,

x 52

= 2514

x 0.4 = 0.75x = 1.15

or

x 52

= 2514

x 0.4 = 0.75x = 0.35

(c) 2x2 + 7x + 2 = 0Solution:

2x2 + 7x + 2 = 0


x2 + 27

x + 1 = 0

x2 + 27

x = 1

x2 + 27

x = 1

x2 + 27

x + 2

47

= 1 + 2

47

2

37x

+ = 1633

Either,

x + 47

= 1633

x + 1.75 = 1.44x = 0.31

or

x + 47

= 1633

x + 1.75 = 1.44x = 3.19

(d) 4x2 = 5x 21Solution:

4x2 = 5x 21

x2 = 45

x 4x

x2 45

x = 421

x2 45

x + 2

85

=

421

+ 2

85

2

85x

= 64311


Either,

x 85

= + 64311

or

x 85

= 64311

Hence, roots of 4x2 = 5x 21 are complex.(c) 2x2 + 5x 3 = 0Solution:

2x2 + 5x 3 = 0

x2 + 25

x 23

= 0

x2 + 25

x = 23

x2 + 25

x + 2

45

=

23

+ 2

45

2

45x

+ = 1649

Either,

x + 45

= 47

x = 42

x = 21

or

x + 45

= 47

x = 412

x = 3(f) 3x2 + 8x + 2 = 0


Solution:3x2 + 8x + 2 = 0

x2 + 38

x + 32

= 0

x2 + 38

x = 32

x2 + 38

x + 2

34

=

32

+ 2

34

2

34x

+ = 9

10

Either,

x + 34

= 9

10

x = 1.054 1.33x = 0.28

or

x + 34

= 9

10

x = 1.054 1.33x = 2:39

(g) 7x2 28x + 15 = 0Solution:

7x2 28x + 15 = 0

x2 4x + 7

15= 0

x2 4x = 715

x2 4x + (2)2 = 715

+ (2)2

(x 2)2 = 7

17

Either,


x 2 = 7

13

x 2 = 1.36x 2 = 3.36

or

x 2 = 7

13

x 2 = 1.36x = 0.64

(h) 5x2 + 12x + 3 = 0Solution:

5x2 + 12x + 3 = 0

x2 + 5

12x +

53

= 0

x2 + 5

12x =

53

x2 + 5

12x +

2

56

=

53

+ 2

56

2

56x

+ = 2521

Either,

x + 56

= 2521

x + 1.10 = 0.918x = 0.918 1.10x = 2.02

(i) 2x2 + 3x 4 = 0Solution:

2x2 + 3x 4 = 0

x2 + 23

x 2 = 0


x2 + 23

x = 2

x2 + 23

x + 2

43

= 2 + 2

43

2

43x

+ = 1641

Either,

x + 43

= 1641

x + 0.75 = 1.6x = 0.85

or

x + 43

= 1641

x + 0.75 = 1.6x = 2.35

(j) x2 12x + 36 = 0Solution:

x2 12x + 36 = 0(x)2 2(x) (6) + (6)2 = 0

(x6)2 = 0(x6) (x6) = 0

x = 6 or 6So, x may be 6 repeated.

(k) 5x2 + 30x 18 = 0 Solution:

5x2 + 30x 18 = 0

x2 + 6x 5

18= 0

x2 + 6x = 5

18


x2 + 6x + (3)2 = 5

18 + (3)2

(x+3)2 = 563

Either,

x + 3 = 563

x + 3 = 3.55x = 0.55

or

x + 3 = 563

x + 3 = 3.55x = 6.55

(l) 3x2 4x + 7 = 0Solution:

3x2 4x + 7 = 0

x2 34

x + 34

= 0

x2 34

x = 47

(x)2 2(x)

32

+2

32

=

37

+ 2

32

2

32x

= 37

+ 94

2

32x

= 9

421 +

2

32x

= 917

Taking square root of each side.


x 32

= 917

Hence, root of 3x2 4x + 7 = 0 are complex.(m) 3x2 + x 2 = 0Solution:

3x2 + x 2 = 0

x2 + 3x

32

= 0

x2 + 3x

= 32

x2 + 3x

+ 2

61

=

32

+ 2

61

2

61x

+ = 3625

Either,

x + 61

= 65

x = 64

x = 32

or

x + 61

= 65

x = 66

x = 1(n) x2 11x 26 = 0Solution:

x2 11x 26 = 0x2 11x = 26


x2 11x + 2

211

= 26 + 2

211

2

211x

= 4

225

Either,

x 2

11=

215

x = 2

26

x = 13or

x 2

11=

215

x = 24

x = 2(o) 3x2 + 5x 2 = 0Solution:

3x2 + 5x 2 = 0

x2 + 35

x 32

= 0

x2 + 35

x = 32

x2 + 35

x + 2

65

=

32

+ 2

65

2

65x

+ = 3649

Either,

x + 65

= 67

x = 62


x = 31

or

x + 65

= 67

x = 612

x = 2(p) 2x 3x2 4 = 0Solution:

2x 2x2 4 = 03x2 2x + 4 = 0

x2 32

x + 34

= 0

x2 32

x = 34

(x)2 2(x)

31

+2

31

=

34

+2

31

2

31x

= 34

+ 91

2

31x

= 9

112 +

2

31x

= 920

Taking square root of each side.

x 31

= 920

Hence, roots of 2x 4 = 0 are complex.(q) x2 7x 30 = 0 Solution:

x2 7x 30 = 0


x2 7x = 20

x2 7x + 2

27

= 30 + 2

27

2

274

= 4

169

Either,

x 27

= 2

13

x = 2

20

x = 10or

x 27

= 213

x = 26

x = 3(r) (2x+3) (x2) x(x+1) = 0Solution:

(2x + 3) (x 2) x (x + 1) = 0(2x2 x 6) (x2 + x) = 0

x2 2x 6 = 0x2 2x = 6

x2 2x + (1)2 = 6 + (1)2 (x 1)2 = 7

Either,x 1 = 7

1 = 2.65x = 3.65

orx = 7

x 1 = 2.65


x = 1.65(s) x2 6x 16 = 0Solution:

x2 6x 16 = 0x2 6x = 16

x2 6x + (3)2 = 16 + (3)2

(x 3)2 = 25Either,

x 3 = 5x = 8

orx 3 = 5

x = 2

(t) 2x2 x 321 = 0

Solution:

2x2 x 321

= 0

2x2 x 27

= 0

x2 2x

27

= 0

x2 2x

= 47

x2 2x

+ 2

41

=

47

+ 2

41

2

41x

= 1629

Either,

x 41

= 1629

x 0.25 = 1.35


x = 1.60or

x 41

= 1629

x 0.25 = 1.35x = 1.10

(u) x2 16x 10 = 0Solution:

x2 16x 10 = 0x2 16x = 10

x2 16x + (8)2 = 10 + (8)2 (x 8)2 = 74

Either,x 8 = 74x 8 = 8.60

x = 16.60or

x 8 = 74x 8 = 8.60

x = 0.60(v) (2x + 1) (5x 4) (3x 2)2 = 0Solution:

(2x + 1) (5x 4) (3x 2)2 = 0(10x2 3x 4) (9x2 12x + 4) = 0

x2 + 9x 8 = 0x2 + 9x = 8

x2 + 9x + 2

29

= 8 + 2

29

2

29x

+ = 4

113

Either,


x + 29

= 4

113

x + 4.5 = 5.32x = 0.82

or

x + 29

= 4

113

x + 4.5 = 5.32x = 9.82

(w) x2 2x 5 = 0Solution:

x2 2x 5 = 0x2 2x = 5

x2 2x + (1)2 = 5 +(1)2 (x 1)2 = 6

Either,x 1 = 6x 1 = 2.45

x =3.45or

x 1 = 6x 1 = 2.45

x = 1.45(x) 5x2 8x 30 = 0Solution:

5x2 8x 30 = 0

x2 5x

6 = 0

x2 5x

x = 6

x2 5x

+ 3

54

= 6+ 3

54


2

54x

= 25

166

Either,

x 54

= 25

166

x 0.8 = 2.58x = 3.38

or

x 54

= 25

166

x 0.8 = 2.58x = 1.78

(y) 4x(3x 1) 2 = (2x 1) (5x + 1)Solution:

4x(3x 1) 2 = (2x 1) (5x + 1)12x2 4x 2 = 10x2 3x 1

12x2 4x 2 + 3x + 1 = 02x2 x 1 = 0

x2 2x

21

= 0

x2 2x

= 21

x2 2x

+ 2

41

=

21

+ 2

41

2

41x

= 169

Either,

x 41

= 43

x = 44

x = 1


or

x 41

= 43

x = 42

x = 21

(z) 5x2 16x + 2 = 0Solution:

5x2 16x + 2 = 0

x2 5

16x +

52

= 0

x2 5

16x =

52

x2 5

16x +

2

58

=

52

+ 2

58

2

58x

= 2554

Either,

x 58

= 2554

x 1.6 = 1.47x = 3.07

or

x 58

= 2554

x 1.6 = 1.47x = 0.13

__________

EXERCISE 1AEXERCISE No. 1BEXERCISE NO.1C

18681858 math o level solution of quadratic equation

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