ML Aggarwal Quadratic Equations Exe-7 Class 9 ICSE Maths APC Understanding Solutions. Solutions of Exercise-7. This post is the Solutions of ML Aggarwal Chapter 7- Quadratic Equations for ICSE Maths Class-9. APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-7. Quadratic Equations for ICSE Board Class-9. Visit official website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Quadratic Equations Exe-7 Class 9 ICSE Maths Solutions
Board | ICSE |
Publications | Avichal Publishig Company (APC) |
Subject | Maths |
Class | 9th |
Chapter-7 | Quadratic Equations |
Writer | ML Aggarwal |
Book Name | Understanding |
Topics | Solution of Exe-7 Questions |
Edition | 2021-2022 |
Exe-7 Solutions of ML Aggarwal for ICSE Class-9 Ch-7, Quadratic Equations
Note:- Before viewing Solutions of Chapter -7 Quadratic Equations Class-9 of ML Aggarwal Solutions . Read the Chapter Carefully. Then solve all example given in Exercise-7, MCQS and Chapter Test.
Quadratic Equations Exe-7
ML Aggarwal Class 9 ICSE Maths Solutions
Page 146
Solve the following (1 to 12) equations:
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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