ML Aggarwal Quadratic Equations Exe-7 Class 9 ICSE Maths Solutions

ML Aggarwal Quadratic Equations Exe-7 Class 9 ICSE Maths Solutions. Step by step solutions of Quadratic Equations as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Quadratic Equations Exe-7 Class 9 ICSE Maths Solutions Ch-7

Board ICSE
Subject Maths
Class 9th
Chapter-7 Quadratic Equations
Topics Solution of Exe-7 Questions
Academic Session 2024-2025

Solution of Quadratic Equations Questions

ML Aggarwal Quadratic Equations Exe-7 Class 9 ICSE Maths Solutions

Question 1. 

(i) x² – 11x + 30 = 0

(ii) 4x² – 25 = 0

Answer:

(i) x² – 11x + 30 = 0

Let us simplify the given equation,

By factorizing, we get

x2 – 5x – 6x + 30 = 0

x(x – 5) – 6 (x – 5) = 0

(x – 5) (x – 6) = 0

So,

(x – 5) = 0 or (x – 6) = 0

x = 5 or x = 6

∴ Value of x = 5, 6

(ii) 4x² – 25 = 0

Let us simplify the given equation,

4x² = 25

x2 = 25/4

x = ± √(25/4)

±5/2

∴ Value of x = +5/2, -5/2

Question 2. 

(i) 2x² – 5x = 0

(ii) x² – 2x = 48

Answer:

(i) 2x² – 5x = 0

Let us simplify the given equation,

x(2x – 5) = 0

so,

x = 0 or 2x – 5 = 0

x = 0 or 2x = 5

x = 0 or x = 5/2

∴ Value of x = 0, 5/2

(ii) x² – 2x = 48

Let us simplify the given equation,

By factorizing, we get

x2 – 2x – 48 = 0

x2 – 8x+ 6x – 48 = 0

x(x – 8) + 6 (x – 8) = 0

(x – 8) (x + 6) = 0

So,

(x – 8) = 0 or (x + 6) = 0

x = 8 or x = -6

∴ Value of x = 8, -6

Question 3. 

(i) 6 + x = x²

(ii) 2x² + 3x + 1= 0

Answer:

(i) 6 + x = x²

Let us simplify the given equation,

6 + x – x2 = 0

x2 – x – 6 = 0

By factorizing, we get

x2 – 3x + 2x – 6 = 0

x(x – 3) + 2 (x – 3) = 0

(x – 3) (x + 2) = 0

So,

(x – 3) = 0 or (x + 2) = 0

x = 3 or x = -2

∴ Value of x = 3, -2

(ii) 2x² + 3x + 1= 0

Let us simplify the given equation,

By factorizing, we get

2x2 – 2x – x + 1 = 0

2x(x – 1) – 1 (x – 1) = 0

(x – 1) (2x – 1) = 0

So,

(x – 1) = 0 or (2x – 1) = 0

x = 1 or 2x = 1

x = 1 or x = ½

∴ Value of x = 1, ½

Question 4. 

(i) 3x² = 2x + 8
(ii) 4x² + 15 = 16x

Answer:

(i) 3x² = 2x + 8

Let us simplify the given equation,

3x2 – 2x – 8 = 0

By factorizing, we get

3x2 – 6x + 4x – 8 = 0

3x(x – 2) + 4 (x – 2) = 0

(x – 2) (3x + 4) = 0

So,

(x – 2) = 0 or (3x + 4) = 0

x = 2 or 3x = -4

x = 2 or x = -4/3

∴ Value of x = 2 or -4/3

(ii) 4x² + 15 = 16x

Let us simplify the given equation,

4x2 – 16x + 15 = 0

By factorizing, we get

4x2 – 6x – 10x + 15 = 0

2x(2x – 3) – 5 (2x – 3) = 0

(2x – 3) (2x – 5) = 0

So,

(2x – 3) = 0 or (2x – 5) = 0

2x = 3 or 2x = 5

x = 3/2 or x = 5/2

∴ Value of x = 3/2 or 5/2

Question 5. 

(i) x (2x + 5) = 25

(ii) (x + 3) (x – 3) = 40

Answer:

(i) x (2x + 5) = 25

Let us simplify the given equation,

2x2 + 5x – 25 = 0

By factorizing, we get

2x2 + 10x – 5x – 25 = 0

2x(x + 5) – 5 (x + 5) = 0

(x + 5) (2x – 5) = 0

So,

(x + 5) = 0 or (2x – 5) = 0

x = -5 or 2x = 5

x = -5 or x = 5/2

∴ Value of x = -5, 5/2

(ii) (x + 3) (x – 3) = 40

Let us simplify the given equation,

x2 – 3x + 3x – 9 = 40

x2 – 9 – 40 = 0

x2 – 49 = 0

x2 = 49

x = √49

± 7

∴ Value of x = 7, -7

Question 6. 

(i) (2x + 3) (x – 4) = 6
(ii) (3x + 1) (2x + 3) = 3

Answer:

(i) (2x + 3) (x – 4) = 6

Let us simplify the given equation,

2x2 – 8x + 3x – 12 – 6 = 0

2x2 – 5x – 18 = 0

By factorizing, we get

2x2 – 9x + 4x – 18 = 0

x (2x – 9) + 2 (2x – 9) = 0

(2x – 9) (x + 2) = 0

So,

(2x – 9) = 0 or (x + 2) = 0

2x = 9 or x = -2

x = 9/2 or x = -2

∴ Value of x = 9/2, -2

(ii) (3x + 1) (2x + 3) = 3

Let us simplify the given equation,

6x2 + 9x + 2x + 3 – 3 = 0

6x2 + 11x = 0

x(6x + 11) = 0

So,

x = 0 or 6x + 11 = 0

x = 0 or 6x = -11

x = 0 or x = -11/6

∴ Value of x = 0, -11/6

Question 7. 

(i) 4x² + 4x + 1 = 0
(ii) (x – 4)² + 5² = 132

Answer:

(i) 4x² + 4x + 1 = 0

Let us simplify the given equation,

By factorizing, we get

4x2 + 2x + 2x + 1 = 0

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

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

So,

(2x + 1) = 0 or (2x + 1) = 0

2x = -1 or 2x = -1

x = -1/2 or x = -1/2

∴ Value of x = -1/2, -1/2

(ii) (x – 4)² + 5² = 132

Let us simplify the given equation,

x2 + 16 – 2(x) (4) + 25 – 169 = 0

x2 – 8x -128 = 0

By factorizing, we get

x2 – 16x + 8x – 128 = 0

x(x – 16) + 8 (x – 16) = 0

(x – 16) (x + 8) = 0

So,

(x – 16) = 0 or (x + 8) = 0

x = 16 or x = -8

∴ Value of x = 16, -8

Question 8. 

(i) 21x2 = 4 (2x + 1)

(ii) 2/3x2 – 1/3x – 1 = 0

Answer

(i) 21x2 = 4 (2x + 1)

Let us simplify the given equation,

21x2 = 8x + 4

21x2 – 8x – 4 = 0

By factorizing, we get

21x2 – 14x + 6x – 4 = 0

7x(3x – 2) + 2(3x – 2) = 0

(3x – 2) (7x + 2) = 0

So,

(3x – 2) = 0 or (7x + 2) = 0

3x = 2 or 7x = -2

x = 2/3 or x = -2/7

∴ Value of x = 2/3 or -2/7

(ii) 2/3x2 – 1/3x – 1 = 0

Let us simplify the given equation,

By taking 3 as LCM and cross multiplying

2x2 – x – 3 = 0

By factorizing, we get

2x2 – 3x + 2x – 3 = 0

x(2x – 3) + 1 (2x – 3) = 0

(2x – 3) (x + 1) = 0

So,

(2x – 3) = 0 or (x + 1) = 0

2x = 3 or x = -1

x = 3/2 or x = -1

∴ Value of x = 3/2, -1

Question 9. 

(i) 6x + 29 = 5/x

(ii) x + 1/x = 2 ½

Answer:

(i) 6x + 29 = 5/x

Let us simplify the given equation,

By cross multiplying, we get

6x2 + 29x – 5 = 0

By factorizing, we get

6x2 + 30x – x – 5 = 0

6x (x + 5) -1 (x + 5) = 0

(x + 5) (6x – 1) = 0

So,

(x + 5) = 0 or (6x – 1) = 0

x = -5 or 6x = 1

x = -5 or x = 1/6

∴ Value of x = -5, 1/6

(ii) x + 1/x = 2 ½

x + 1/x = 5/2

Let us simplify the given equation,

By taking LCM

x2 + 1 = 5x/2

By cross multiplying,

2x2 + 2 – 5x = 0

2x2 – 5x + 2 = 0

By factorizing, we get

2x2 – x – 4x + 2 = 0

x(2x – 1) – 2 (2x – 1) = 0

(2x – 1) (x – 2) = 0

So,

(2x – 1) = 0 or (x – 2) = 0

2x = 1 or x = 2

x = ½ or x = 2

∴ Value of x = ½, 2

Question 10. 

(i) 3x – 8/x = 2

(ii) x/3 + 9/x = 4

Answer:

(i) 3x – 8/x = 2

Let us simplify the given equation,

By taking LCM and cross multiplying,

3x2 – 8 = 2x

3x2 – 2x – 8 = 0

By factorizing, we get

3x2 – 6x + 4x – 8 = 0

3x(x – 2) + 4 (x – 2) = 0

(x – 2) (3x + 4) = 0

So,

(x – 2) = 0 or (3x + 4) = 0

x = 2 or 3x = -4

x = 2 or x = -4/3

∴ Value of x = 2, -4/3

(ii) x/3 + 9/x = 4

Let us simplify the given equation,

By taking 3x as LCM and cross multiplying

x2 + 27 = 12x

x2 – 12x + 27 = 0

By factorizing, we get

x2 – 3x – 9x + 27 = 0

x (x – 3) – 9 (x – 3) = 0

(x – 3) (x – 9) = 0

So,

(x – 3) = 0 or (x – 9) = 0

x = 3 or x = 9

∴ Value of x = 3, 9

Question 11. 

(i) (x – 1)/(x + 1) = (2x – 5)/(3x – 7)

(ii) 1/(x + 2) + 1/x = ¾

Answer:

(i) (x – 1)/(x + 1) = (2x – 5)/(3x – 7)

Let us simplify the given equation,

By cross multiplying,

(x – 1) (3x – 7) = (2x – 5) (x + 1)

3x2 – 7x – 3x + 7 = 2x2 + 2x – 5x – 5

3x2 – 10x + 7 – 2x2 +3x + 5 = 0

x2 – 7x + 12 = 0

By factorizing, we get

x2 – 4x – 3x + 12 = 0

x (x – 4) – 3 (x – 4) = 0

(x – 4) (x – 3) = 0

So,

(x – 4) = 0 or (x – 3) = 0

x = 4 or x = 3

∴ Value of x = 4, 3

(ii) 1/(x + 2) + 1/x = ¾

Let us simplify the given equation,

By taking x(x + 2) as LCM

(x+x+2)/x(x + 2) = ¾

By cross multiplying,

4(2x + 2) = 3x(x + 2)

8x + 8= 3x2 + 6x

3x2 + 6x – 8x – 8 = 0

3x2 – 2x – 8 = 0

By factorizing, we get

3x2 – 6x + 4x – 8 = 0

3x(x – 2) + 4 (x – 2) = 0

(x – 2) (3x + 4) = 0

So,

(x – 2) = 0 or (3x + 4) = 0

x = 2 or 3x = -4

x = 2 or x = -4/3

∴ Value of x = 2, -4/3

Question 12. 

(i) 8/(x + 3) – 3/(2 – x) = 2

(ii) x/(x + 1) + (x + 1)/x = 2 1/6

Answer:

(i) 8/(x + 3) – 3/(2 – x) = 2

Let us simplify the given equation,

By taking (x+3)(2-x) as LCM

[8(2-x) – 3(x+3)] / (x+3) (2-x) = 2

[16 – 8x – 3x – 9] / [2x – x2 + 6 – 3x] = 2

[-11x + 7] = 2(-x2 – x + 6)

7 – 11x= -2x2 – 2x + 12

2x2 + 2x – 11 x – 12 + 7 = 0

2x2 – 9x – 5 = 0

By factorizing, we get

2x2 – 10x + x – 5 = 0

2x (x – 5) + 1 (x – 5) = 0

(x – 5) (2x + 1) = 0

So,

(x – 5) = 0 or (2x + 1) = 0

x = 5 or 2x= -1

x = 5 or x = -1/2

∴ Value of x = 5, -1/2

(ii) x/(x + 1) + (x + 1)/x = 2 1/6

x/(x + 1) + (x + 1)/x = 13/6

Let us simplify the given equation,

By taking x(x+1) as LCM

[x(x) + (x+1) (x+1)] / x(x + 1) = 13/6

6[x2 + x2 + x + x + 1] = 13x(x + 1)

6[2x2 + 2x + 1] = 13x2 + 13x

12x2 + 12x + 6 – 13x2 – 13x = 0

-x2 – x + 6 = 0

x2 + x – 6 = 0

By factorizing, we get

x2 + 3x – 2x – 6 = 0

x (x + 3) – 2 (x + 3) = 0

(x + 3) (x – 2) = 0

So,

(x + 3) = 0 or (x – 2) = 0

x = -3 or x = 2

∴ Value of x = -3, 2

—  : End of ML Aggarwal Quadratic Equations Exe – 7 Class 9 ICSE Maths Solutions :–

Return to :- ML Aggarawal Maths Solutions for ICSE  Class-9

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