ML Aggarwal Quadratic Equations Chapter Test 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 Chapter Test Class 9 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-7 | Quadratic Equations |
| Topics | Solution of Ch-Test Questions |
| Academic Session | 2024-2025 |
Solution of Ch-Test Questions on Quadratic Equations
ML Aggarwal Quadratic Equations Chapter Test Class 9 ICSE Maths Solutions
Question 1.
(i) x (2x + 5) = 3
(ii) 3x2 – 4x – 4 = 0.
Answer:
(i) x (2x + 5) = 3
We can write it as
2x2 + 5x – 3 = 0
By further calculation
2x2 + 6x – x – 3 = 0
By taking out the common terms
2x (x + 3) – 1 (x + 3) = 0
So we get
(x + 3) (2x – 1) = 0
Here
x + 3 = 0 then x = – 3
2x – 1 = 0 then 2x = 1 where x = ½
Therefore, x = – 3, ½.
(ii) 3x2 – 4x – 4 = 0
We can write it as
3x2 – 6x + 2x – 4 = 0
By taking out the common terms
3x (x – 2) + 2 (x – 2) = 0
So we get
(x – 2) (3x + 2) = 0
Here
x – 2 = 0 then x = 2
3x + 2 = 0 then 3x = – 2 where x = – 2/3
Therefore, x = 2, – 2/3.
Question 2.
(i) 4x2 – 2x + ¼ = 0
(ii) 2x2 + 7x + 6 = 0.
Answer:
(i) 4x2 – 2x + ¼ = 0
Multiply the equation by 4
16x2 – 8x + 1 = 0
We can write it as
16x2 – 4x – 4x + 1 = 0
Taking out the common terms
4x (4x – 1) – 1 (4x – 1) = 0
So we get
(4x – 1) (4x – 1) = 0
(4x – 1)2 = 0
Here
4x – 1 = 0
4x = 1
By division
x = ¼, ¼
(ii) 2x2 + 7x + 6 = 0
We can write it as
2x2 + 4x + 3x + 6 = 0
By further calculation
2x (x + 2) + 3 (x + 2) = 0
So we get
(x + 2) (2x + 3) = 0
Here
x + 2 = 0 then x = – 2
2x + 3 = 0 then 2x = – 3 where x = – 3/2
x = – 2, – 3/2
Question 3.
(i) (x – 1)/ (x – 2) + (x – 3)/ (x – 4) = 3 1/3
(ii) 6/x – 2/(x – 1) = 1/(x – 2).
Answer:
(i) (x – 1)/ (x – 2) + (x – 3)/ (x – 4) = 3 1/3
By taking LCM
[(x – 1) (x – 4) + (x – 2) (x – 3)]/ (x – 2) (x – 4) = 10/3
By further calculation
(x2 – 5x + 4 + x2 – 5x + 6)/ (x2 – 6x + 8) = 10/3
So we get
(2x2 – 10x + 10)/ (x2 – 6x + 8) = 10/3
By cross multiplication
10x2 – 60x + 80 = 6x2 – 30x + 30
By further simplification
10x2 – 60x + 80 – 6x2 + 30x – 30 = 0
So we get
4x2 – 30x + 50 = 0
Dividing by 2
2x2 – 15x + 25 = 0
It can be written as
2x2 – 10x – 5x + 25 = 0
Taking out the common terms
2x (x – 5) – 5 (x – 5) = 0
(x – 5) (2x – 5) = 0
Here
x – 5 = 0 then x = 5
2x – 5 = 0 then 2x = 5 where x = 5/2
Therefore, x = 5, 5/2.
(ii) 6/x – 2/(x – 1) = 1/(x – 2)
Taking LCM
(6x – 6 – 2x)/ x (x – 1) = 1/ (x – 2)
By further calculation
(4x – 6)/ (x2 – x) = 1/(x – 2)
By cross multiplication
4x2 – 8x – 6x + 12 = x2 – x
So we get
4x2 – 14x + 12 – x2 + x = 0
3x2 – 13x + 12 = 0
3x2 – 4x – 9x + 12 = 0
Taking out the common terms
x (3x – 4) – 3 (3x – 4) = 0
(3x – 4) (x – 3) = 0
Here
3x – 4 = 0 then 3x = 4 where x = 4/3
x – 3 = 0 then x = 3
Therefore, x = 3, 4/3.
— : End of ML Aggarwal Quadratic Equations Chapter Test Class 9 ICSE Maths Solutions :–
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