Selina Textbook Solutions: Quadratic Equations | Mathematics Class 10 ICSE PDF Download

 Page 1


Quadratic Equations 
 
 
Question 1.  
Find which of the following equations are quadratic: 
Solution 1(i) 
(3x – 1)
2
 = 5(x + 8) 
? (9x
2
 – 6x + 1) = 5x + 40 
? 9x
2
 – 11x – 39 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(ii) 
5x
2
 – 8x = -3(7 – 2x) 
? 5x
2
 – 8x = 6x – 21 
? 5x
2
 – 14x + 21 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(iii) 
(x – 4)(3x + 1) = (3x – 1)(x +2) 
? 3x
2
 + x – 12x – 4 = 3x
2
 + 6x – x – 2 
? 16x + 2 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(iv) 
x
2
 + 5x – 5 = (x – 3)
2 
? x
2
 + 5x – 5 = x
2
 – 6x + 9 
? 11x – 14 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(v) 
7x
3
 – 2x
2
 + 10 = (2x – 5)
2 
? 7x
3
 – 2x
2
 + 10 = 4x
2
 – 20x + 25 
? 7x
3
 – 6x
2
 + 20x – 15 = 0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(vi) 
(x – 1)
2
 + (x + 2)
2
 + 3(x +1) = 0 
? x
2
 – 2x + 1 + x
2
 + 4x + 4 + 3x + 3 = 0 
? 2x
2
 + 5x + 8 = 0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Page 2


Quadratic Equations 
 
 
Question 1.  
Find which of the following equations are quadratic: 
Solution 1(i) 
(3x – 1)
2
 = 5(x + 8) 
? (9x
2
 – 6x + 1) = 5x + 40 
? 9x
2
 – 11x – 39 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(ii) 
5x
2
 – 8x = -3(7 – 2x) 
? 5x
2
 – 8x = 6x – 21 
? 5x
2
 – 14x + 21 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(iii) 
(x – 4)(3x + 1) = (3x – 1)(x +2) 
? 3x
2
 + x – 12x – 4 = 3x
2
 + 6x – x – 2 
? 16x + 2 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(iv) 
x
2
 + 5x – 5 = (x – 3)
2 
? x
2
 + 5x – 5 = x
2
 – 6x + 9 
? 11x – 14 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(v) 
7x
3
 – 2x
2
 + 10 = (2x – 5)
2 
? 7x
3
 – 2x
2
 + 10 = 4x
2
 – 20x + 25 
? 7x
3
 – 6x
2
 + 20x – 15 = 0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(vi) 
(x – 1)
2
 + (x + 2)
2
 + 3(x +1) = 0 
? x
2
 – 2x + 1 + x
2
 + 4x + 4 + 3x + 3 = 0 
? 2x
2
 + 5x + 8 = 0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Question 2(i) 
Is x = 5 a solution of the quadratic equation x
2
 – 2x – 15 = 0? 
 
Solution: 
x
2
 – 2x – 15 = 0 
For x = 5 to be solution of the given quadratic equation it should satisfy the equation. 
So, substituting x = 5 in the given equation, we get 
L.H.S = (5)
2
 – 2(5) – 15 
= 25 – 10 – 15 
= 0 
= R.H.S 
Hence, x = 5 is a solution of the quadratic equation x
2
 – 2x – 15 = 0. 
Question 2(ii). 
Is x = -3 a solution of the quadratic equation 2x
2
 – 7x + 9 = 0? 
 
Solution: 
2x
2
 – 7x + 9 = 0 
For x = -3 to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x = 5 in the given equation, we get 
L.H.S =2(-3)
2
 – 7(-3) + 9 
= 18 + 21 + 9 
= 48 
? R.H.S 
Hence, x = -3 is not a solution of the quadratic equation 2x
2
 – 7x + 9 = 0. 
Question 3. 
If  is a solution of equation 3x
2
 + mx + 2 = 0, find the value of m. 
 
Solution: 
For x =  to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x =  in the given equation, we get 
 
Page 3


Quadratic Equations 
 
 
Question 1.  
Find which of the following equations are quadratic: 
Solution 1(i) 
(3x – 1)
2
 = 5(x + 8) 
? (9x
2
 – 6x + 1) = 5x + 40 
? 9x
2
 – 11x – 39 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(ii) 
5x
2
 – 8x = -3(7 – 2x) 
? 5x
2
 – 8x = 6x – 21 
? 5x
2
 – 14x + 21 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(iii) 
(x – 4)(3x + 1) = (3x – 1)(x +2) 
? 3x
2
 + x – 12x – 4 = 3x
2
 + 6x – x – 2 
? 16x + 2 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(iv) 
x
2
 + 5x – 5 = (x – 3)
2 
? x
2
 + 5x – 5 = x
2
 – 6x + 9 
? 11x – 14 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(v) 
7x
3
 – 2x
2
 + 10 = (2x – 5)
2 
? 7x
3
 – 2x
2
 + 10 = 4x
2
 – 20x + 25 
? 7x
3
 – 6x
2
 + 20x – 15 = 0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(vi) 
(x – 1)
2
 + (x + 2)
2
 + 3(x +1) = 0 
? x
2
 – 2x + 1 + x
2
 + 4x + 4 + 3x + 3 = 0 
? 2x
2
 + 5x + 8 = 0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Question 2(i) 
Is x = 5 a solution of the quadratic equation x
2
 – 2x – 15 = 0? 
 
Solution: 
x
2
 – 2x – 15 = 0 
For x = 5 to be solution of the given quadratic equation it should satisfy the equation. 
So, substituting x = 5 in the given equation, we get 
L.H.S = (5)
2
 – 2(5) – 15 
= 25 – 10 – 15 
= 0 
= R.H.S 
Hence, x = 5 is a solution of the quadratic equation x
2
 – 2x – 15 = 0. 
Question 2(ii). 
Is x = -3 a solution of the quadratic equation 2x
2
 – 7x + 9 = 0? 
 
Solution: 
2x
2
 – 7x + 9 = 0 
For x = -3 to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x = 5 in the given equation, we get 
L.H.S =2(-3)
2
 – 7(-3) + 9 
= 18 + 21 + 9 
= 48 
? R.H.S 
Hence, x = -3 is not a solution of the quadratic equation 2x
2
 – 7x + 9 = 0. 
Question 3. 
If  is a solution of equation 3x
2
 + mx + 2 = 0, find the value of m. 
 
Solution: 
For x =  to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x =  in the given equation, we get 
 
Question 4. 
 and 1 are the solutions of equation mx
2
 + nx + 6 = 0. Find the values of m and n. 
 
Solution: 
 
For x =   and x = 1 to be solutions of the given quadratic equation it should satisfy the 
equation 
 
So, substituting x =   and x = 1 in the given equation, we get 
 
 
 
Solving equations (1) and (2) simultaneously, 
4m  + 6n + 54 = 0 …..(1) 
m + n  + 6 = 0 ….(2) 
(1) – (2) × 6 
? -2m + 18 = 0 
? m = 9 
Substitute in (2) 
? n = -15 
Question 5. 
If 3 and -3 are the solutions of equation ax
2
 + bx – 9 = 0. Find the values of a and b. 
 
Solution: 
 
For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the 
equation 
 
So, substituting x = 3 and x = -3 in the given equation, we get 
 
 
Solving equations (1) and (2) simultaneously, 
Page 4


Quadratic Equations 
 
 
Question 1.  
Find which of the following equations are quadratic: 
Solution 1(i) 
(3x – 1)
2
 = 5(x + 8) 
? (9x
2
 – 6x + 1) = 5x + 40 
? 9x
2
 – 11x – 39 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(ii) 
5x
2
 – 8x = -3(7 – 2x) 
? 5x
2
 – 8x = 6x – 21 
? 5x
2
 – 14x + 21 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(iii) 
(x – 4)(3x + 1) = (3x – 1)(x +2) 
? 3x
2
 + x – 12x – 4 = 3x
2
 + 6x – x – 2 
? 16x + 2 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(iv) 
x
2
 + 5x – 5 = (x – 3)
2 
? x
2
 + 5x – 5 = x
2
 – 6x + 9 
? 11x – 14 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(v) 
7x
3
 – 2x
2
 + 10 = (2x – 5)
2 
? 7x
3
 – 2x
2
 + 10 = 4x
2
 – 20x + 25 
? 7x
3
 – 6x
2
 + 20x – 15 = 0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(vi) 
(x – 1)
2
 + (x + 2)
2
 + 3(x +1) = 0 
? x
2
 – 2x + 1 + x
2
 + 4x + 4 + 3x + 3 = 0 
? 2x
2
 + 5x + 8 = 0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Question 2(i) 
Is x = 5 a solution of the quadratic equation x
2
 – 2x – 15 = 0? 
 
Solution: 
x
2
 – 2x – 15 = 0 
For x = 5 to be solution of the given quadratic equation it should satisfy the equation. 
So, substituting x = 5 in the given equation, we get 
L.H.S = (5)
2
 – 2(5) – 15 
= 25 – 10 – 15 
= 0 
= R.H.S 
Hence, x = 5 is a solution of the quadratic equation x
2
 – 2x – 15 = 0. 
Question 2(ii). 
Is x = -3 a solution of the quadratic equation 2x
2
 – 7x + 9 = 0? 
 
Solution: 
2x
2
 – 7x + 9 = 0 
For x = -3 to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x = 5 in the given equation, we get 
L.H.S =2(-3)
2
 – 7(-3) + 9 
= 18 + 21 + 9 
= 48 
? R.H.S 
Hence, x = -3 is not a solution of the quadratic equation 2x
2
 – 7x + 9 = 0. 
Question 3. 
If  is a solution of equation 3x
2
 + mx + 2 = 0, find the value of m. 
 
Solution: 
For x =  to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x =  in the given equation, we get 
 
Question 4. 
 and 1 are the solutions of equation mx
2
 + nx + 6 = 0. Find the values of m and n. 
 
Solution: 
 
For x =   and x = 1 to be solutions of the given quadratic equation it should satisfy the 
equation 
 
So, substituting x =   and x = 1 in the given equation, we get 
 
 
 
Solving equations (1) and (2) simultaneously, 
4m  + 6n + 54 = 0 …..(1) 
m + n  + 6 = 0 ….(2) 
(1) – (2) × 6 
? -2m + 18 = 0 
? m = 9 
Substitute in (2) 
? n = -15 
Question 5. 
If 3 and -3 are the solutions of equation ax
2
 + bx – 9 = 0. Find the values of a and b. 
 
Solution: 
 
For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the 
equation 
 
So, substituting x = 3 and x = -3 in the given equation, we get 
 
 
Solving equations (1) and (2) simultaneously, 
9a + 3b – 9 = 0 …(1) 
9a – 3b – 9 = 0 …(2) 
(1) + (2) 
? 18a – 18 = 0 
? a = 1 
Substitute in (2) 
? b = 0 
Exercise 5B 
Question 1. 
Without solving, comment upon the nature of roots of each of the following equations : 
(i) 7x
2
 – 9x +2 =0 
(ii) 6x
2
 – 13x +4 =0 
(iii) 25x
2
 – 10x +1=0 
(iv) x
2
 + 2v3x – 9=0 
(v) x
2
 – ax – b
2
 =0 
(vi) 2x
2
 +8x +9=0 
 
Solution: 
 
 
Page 5


Quadratic Equations 
 
 
Question 1.  
Find which of the following equations are quadratic: 
Solution 1(i) 
(3x – 1)
2
 = 5(x + 8) 
? (9x
2
 – 6x + 1) = 5x + 40 
? 9x
2
 – 11x – 39 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(ii) 
5x
2
 – 8x = -3(7 – 2x) 
? 5x
2
 – 8x = 6x – 21 
? 5x
2
 – 14x + 21 =0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Solution 1(iii) 
(x – 4)(3x + 1) = (3x – 1)(x +2) 
? 3x
2
 + x – 12x – 4 = 3x
2
 + 6x – x – 2 
? 16x + 2 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(iv) 
x
2
 + 5x – 5 = (x – 3)
2 
? x
2
 + 5x – 5 = x
2
 – 6x + 9 
? 11x – 14 =0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(v) 
7x
3
 – 2x
2
 + 10 = (2x – 5)
2 
? 7x
3
 – 2x
2
 + 10 = 4x
2
 – 20x + 25 
? 7x
3
 – 6x
2
 + 20x – 15 = 0; which is not of the form ax
2
 + bx + c = 0. 
? Given equation is not a quadratic equation. 
Solution 1(vi) 
(x – 1)
2
 + (x + 2)
2
 + 3(x +1) = 0 
? x
2
 – 2x + 1 + x
2
 + 4x + 4 + 3x + 3 = 0 
? 2x
2
 + 5x + 8 = 0; which is of the form ax
2
 + bx + c = 0. 
? Given equation is a quadratic equation. 
Question 2(i) 
Is x = 5 a solution of the quadratic equation x
2
 – 2x – 15 = 0? 
 
Solution: 
x
2
 – 2x – 15 = 0 
For x = 5 to be solution of the given quadratic equation it should satisfy the equation. 
So, substituting x = 5 in the given equation, we get 
L.H.S = (5)
2
 – 2(5) – 15 
= 25 – 10 – 15 
= 0 
= R.H.S 
Hence, x = 5 is a solution of the quadratic equation x
2
 – 2x – 15 = 0. 
Question 2(ii). 
Is x = -3 a solution of the quadratic equation 2x
2
 – 7x + 9 = 0? 
 
Solution: 
2x
2
 – 7x + 9 = 0 
For x = -3 to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x = 5 in the given equation, we get 
L.H.S =2(-3)
2
 – 7(-3) + 9 
= 18 + 21 + 9 
= 48 
? R.H.S 
Hence, x = -3 is not a solution of the quadratic equation 2x
2
 – 7x + 9 = 0. 
Question 3. 
If  is a solution of equation 3x
2
 + mx + 2 = 0, find the value of m. 
 
Solution: 
For x =  to be solution of the given quadratic equation it should satisfy the equation 
So, substituting x =  in the given equation, we get 
 
Question 4. 
 and 1 are the solutions of equation mx
2
 + nx + 6 = 0. Find the values of m and n. 
 
Solution: 
 
For x =   and x = 1 to be solutions of the given quadratic equation it should satisfy the 
equation 
 
So, substituting x =   and x = 1 in the given equation, we get 
 
 
 
Solving equations (1) and (2) simultaneously, 
4m  + 6n + 54 = 0 …..(1) 
m + n  + 6 = 0 ….(2) 
(1) – (2) × 6 
? -2m + 18 = 0 
? m = 9 
Substitute in (2) 
? n = -15 
Question 5. 
If 3 and -3 are the solutions of equation ax
2
 + bx – 9 = 0. Find the values of a and b. 
 
Solution: 
 
For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the 
equation 
 
So, substituting x = 3 and x = -3 in the given equation, we get 
 
 
Solving equations (1) and (2) simultaneously, 
9a + 3b – 9 = 0 …(1) 
9a – 3b – 9 = 0 …(2) 
(1) + (2) 
? 18a – 18 = 0 
? a = 1 
Substitute in (2) 
? b = 0 
Exercise 5B 
Question 1. 
Without solving, comment upon the nature of roots of each of the following equations : 
(i) 7x
2
 – 9x +2 =0 
(ii) 6x
2
 – 13x +4 =0 
(iii) 25x
2
 – 10x +1=0 
(iv) x
2
 + 2v3x – 9=0 
(v) x
2
 – ax – b
2
 =0 
(vi) 2x
2
 +8x +9=0 
 
Solution: 
 
 
 
 
Question 2. 
Find the value of p, if the following quadratic equation has equal roots : 4x
2
 – (p – 2)x + 
1 = 0 
 
Solution: