ML Aggarwal Quadratic Equations Exe-5.3 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-5.3 Questions ( Using formula ) for Quadratic Equations in One Variable as council prescribe guideline for upcoming board exam.  Visit official Website CISCE  for detail information about ICSE Board Class-10.

ML Aggarwal Ch-5 Quadratic Equations in one Variable Exercise- 5.3 Class 10 ICSE Maths Solutions

Question -1

(i) 2x² – 7x + 6 = 0
(ii) 2x² – 6x + 3 =0

Answer -1

let us consider,

a=2,b=-7,c=6

so ,by using the formula

value of X = 2,OR 3/2

Question -2

(i) 256x² – 32x + 1 = 0
(ii) 25x² + 30x + 7 = 0

Answer-

Question-3

(i) 2x² + √5x – 5 = 0
(ii) √3x² + 10x – 8√3 = 0

Answer -3

(i) 2x² + √5x – 5 = 0
Here a = 2, b = √5, c = -5
using the formula

D = b² – 4ac

= (√5)² – 4(2) (-5)

= 5 + 40

= 45

x = [-(√5) ± √45] / 2(2)

= [-√5 ± 3√5)]/ 4

= [-√5 + 3√5)]/ 4 or [-√5 – 3√5]/ 4

= 2√5/4 or -4√5/4

= √5/2 or -√5

Let us consider,

a = √3, b = 10, c = -8√3

using the formula,

D = b² – 4ac

= (10)² – 4(√3) (-8√3)

= 100 + 96

= 196

x = [-(10) ± √196] / 2(√3)

= [-10 ± 14] / 2(√3)

= [-10 + 14)]/ 2√3 or [-10 – 14)]/ 2√3

= 4/2√3 or -24/2√3

Question-4

(i)( x-2)/(x+2)+(x+2)/( x-2)=4
(ii)( x +1)/(x+3)=(3x+2)/(2x+3)

Answer-4

Question-5

(i) a (x² + 1) = (a² + 1) x , a ≠ 0
(ii) 4x² – 4ax + (a² – b²) = 0

Answer-5

(i) a (x² + 1) = (a² + 1) x
ax² – (a² + 1)x + a = 0

Let us consider,

a = a, b = -(a² + 1), c = a

using the formula,

D = b² – 4ac

= (-(a² + 1))² – 4(a) (a)

= a⁴ + 2a² + 1 – 4a²

= a⁴ – 2a² + 1

= (a² – 1)²

putting value of D in  formula

x = [-(-(a² + 1)) ± √(a² – 1)²] / 2(a)

= [(a² + 1) ± (a² – 1)] / 2a

= [(a² + 1) + (a² – 1)] / 2a or [(a² + 1) – (a² – 1)] / 2a

= [a² + 1 + a² – 1]/2a or [a² + 1 – a² + 1]/2a

= 2a²/2a or 2/2a

= a or 1/a

∴ Value of x = a, 1/a

(ii) 4x² – 4ax + (a² – b²) = 0

Let us consider,

a = 4, b = -4a, c = (a² – b²)

using the formula,

D = b² – 4ac

= (-4a)² – 4(4) (a² – b²)

= 16a² – 16(a² – b²)

= 16a² – 16a² + 16b²

= 16b²

putting value of D in  formula

x = [-(-4a) ± √16b²] / 2(4)

= [4a ± 4b] / 8

= 4[a ± b] / 8

= [a ± b] / 2

= [a + b] / 2 or [a – b] / 2

∴ Value of x = [a + b] / 2, [a – b] / 2

Question-6

(i)x−1/x = 3, x ≠ 0
(ii)1/ x  +  1/(x−2) = 3, x ≠ 0, 2

Answer-6

(i) x−1/x = 3, x ≠ 0

Let us simplify the given expression,

By taking LCM

x² – 1 = 3x

x² – 3x – 1 = 0

Let us consider,

a = 1, b = -3, c = -1

using the formula,

D = b² – 4ac

= (-3)² – 4(1) (-1)

= 9 + 4

= 13

Putting Value of D in Formula

x = [-(-3) ± √13] / 2(1)

= [3 ± √13] / 2

= [3 + √13] / 2 or [3 – √13] / 2

∴ Value of x = [3 + √13] / 2 or [3 – √13] / 2

(ii) 1/ x  +  1/(x−2) = 3, x ≠ 0, 2

[(x – 2) + x] / [x(x – 2)] = 3

[x – 2 + x] / [x² – 2x] = 3

2x – 2 = 3(x² – 2x)

2x – 2 = 3x² – 6x

3x² – 6x – 2x + 2 = 0

3x² – 8x + 2

a = 3, b = -8, c = 2

using the formula,

D = b² – 4ac

= (-8)² – 4(3) (2)

= 64 – 24

= 40

Putting value of D in formula

x = [-(-8) ± √40] / 2(3)

= [8 ± 2√10] / 6

= 2[4 ± √10] / 6

= [4 ± √10] / 3

= [4 + √10] / 3 or [4 – √10] / 3

∴ Value of x = [4 + √10] / 3 or [4 – √10] / 3

Question-7 Solve for x

 

Answer-7

Let (2x-1 )/(x+3) = x

then (x+3)/ (2x-1)  = 1/x

2x – 3/x = 5

2x² – 3 = 5x

2x² – 5x – 3 = 0

a = 2, b = -5, c = -3

using the formula,

D = b² – 4ac

= (-5)² – 4(2) (-3)

= 25 + 24

= 49

Putting value of D in equation

x = [-(-5) ± √49] / 2(2)

= [5 ± 7] / 4

= [5 + 7] / 4 or [5 – 7] / 4

= [12]/4 or [-2]/4

= 3 or -1/2

So, x = 3 or -1/2

When x = 3, then

(2x-1) / (x+3) =  3

2x – 1 = 3x + 9

3x + 9 – 2x + 1 = 0

x + 10 = 0

x = -10

When x = -1/2, then

(2x-1) / (x+3) =  -1/2

2(2x – 1) = -(x + 3)

4x – 2 = -x – 3

4x – 2 + x + 3 = 0

5x + 1 = 0

5x = -1

x = -1/5

∴ Value of x = -10, -1/5

Question -8 Solve the following equation by using quadratic equations for x and give your
(i) x² – 5x – 10 = 0
(ii) x² + 7x  = 7

Answer -8

(i) x² – 5x – 10 = 0
On comparing with, ax² + bx + c = 0

a = 1, b = -5, c = -10

using the formula,

D = b² – 4ac

= (-5)² – 4(1) (-10)

= 25 + 40

= 65

putting value of D in equation

x = [-(-5) ± √65] / 2(1)

= [5 ± √65] / 2

= [5 ± 8.06] / 2

= [5 + 8.06] / 2 or [5 – 8.06] / 2

= [13.06]/2 or [-3.06]/2

= 6.53 or -1.53

∴ Value of x = 6.53 or -1.53

(ii) x² + 7x = 7

x² + 7x – 7 = 0

Let us consider,

a = 1, b = 7, c = -7

using the formula,

D = b² – 4ac

= (7)² – 4(1) (-7)

= 49 + 28

= 77

Putting value of D in equation

x = [-7 ± √77] / 2(1)

= [-7 ± 8.77] / 2

= [-7 + 8.77] / 2 or [-7 – 8.77] / 2

= 1.77/2 or -15.77/2

= 0.885 or -7.885

Question-9 Solve the following equations by using quadratic formula and give your answer correct to 2 decimal places :
(i) 4x² – 5x – 3 = 0
(ii) 2x – 1/x = 7

Answer-9

(i) Given equation 4x² – 5x – 3 = 0
Comparing with ax² + bx + c = 0, we have

a = 4, b = -5, c = -3

using the formula,

D = b² – 4ac

= (-5)² – 4(4) (-3)

= 25 + 48

= 73

Putting value of D in equation

x = [-(-5) ± √73] / 2(4)

= [5 ± 8.54] / 8

= [5 + 8.54] / 8 or [5 – 8.54] / 8

= 13.54/8 or -3.54/8

= 1.6925 or -0.4425

∴ Value of x = 1.69 or -0.44

(ii) 2x – 1/x = 7

2x² – 1 = 7x

2x² – 7x – 1 = 0

a = 2, b = -7, c = -1

using the formula,

D = b² – 4ac

= (-7)² – 4(2) (-1)

= 49 + 8

= 57

Putting value of D in equation

x = [-(-7) ± √57] / 2(2)

= [7 ± 7.549] / 4

= [7 + 7.549] / 4 or [7 – 7.549] / 4

= 14.549/4 or -0.549/4

= 3.637 or -0.137

= 3.64 or -0.14

Question-10  Solve for x the quadratic equation Give your answer correct to three significant figures.

(i ) x² – 4x – 8 = 0.  

(ii) x – 18/x = 6. 

Answer-10

(i) Given equation:

x² – 4x – 8 = 0

on comparing

a = 1, b = -4, c = -8

using the formula,

D = b² – 4ac

= (-4)² – 4(1) (-8)

= 16 + 32

= 48

Putting value of D in equation

x = [-(-4) ± √48] / 2(1)

= [4 ± 6.93]/2

= [4 + 6.93]/2 or [4 – 6.93]/2

= [10.93]/2 or -2.93/2

= 5.465 or -1.465

(ii)

x−18/x = 6

x² – 18 = 6x

x² – 6x – 18 = 0

a = 1, b = -6, c = -18

using the formula,

D = b² – 4ac

= (-6)² – 4(1) (-18)

= 36 + 72

= 108

putting value of D in equation

x = [-(-6) ± √108] / 2(1)

= [6 ± 10.39]/2

= [6 + 10.39]/2 or [6 – 10.39]/2

= [16.39]/2 or -4.39/2

= 8.19 or -2.19

Question- 11 Solve the equation 5x² – 3x – 4 = 0 and give your answer correct to 3 significant figures:

Answer-11

We have 5x² – 3x – 4 = 0
Here a = 5, b = – 3, c = – 4

using the formula,

D = b² – 4ac

= (-3)² – 4(5) (-4)

= 9 + 80

= 89

Putting value of D in equation

x = [-(-3) ± √89] / 2(5)

= [3 ± 9.43] / 10

= [3 + 9.43] / 10 or [3 – 9.43] / 10

= 12.433/10 or -6.43/10

= 1.24 or -0.643

— : End of ML Aggarwal Quadratic Equations Exe-5.3 Class 10 ICSE Maths Solutions :–

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