
ICSE Class 10 Maths Selina Solutions Chapter 5 Quadratic Equations PDF
1. Find which of the following equations are quadratic:
(i) (3x – 1) 2 = 5(x + 8)
(ii) 5x 2 – 8x = -3(7 – 2x)
(iii) (x -4) (3x + 1) = (3x – 1) (x + 2)
(iv) x 2 + 5x – 5 = (x – 3) 2
(v) 7x 3 – 2x 2 + 10 = (2x – 5) 2
(vi) (x – 1) 2 + (x + 2) 2 + 3(x + 1) = 0
Solution:
(i) (3x – 1) 2 = 5(x + 8) ⇒ (9x 2 – 6x + 1) = 5x + 40 ⇒ 9x 2 – 11x – 39 = 0; which is of the general form ax 2 + bx + c = 0. Thus, the given equation is a quadratic equation.2. (i) Is x = 5 a solution of the quadratic equation x 2 – 2x – 15 = 0?
Solution:
Given quadratic equation, x 2 – 2x – 15 = 0 We know that, for x = 5 to be a solution of the given quadratic equation it should satisfy the equation. Now, on substituting x = 5 in the given equation, we have L.H.S = (5) 2 – 2(5) – 15 = 25 – 10 – 15 = 0 = R.H.S Therefore, x = 5 is a solution of the given quadratic equation x 2 – 2x – 15 = 0(ii) Is x = -3 a solution of the quadratic equation 2x 2 – 7x + 9 = 0?
Solution:
Given quadratic equation, 2x 2 – 7x + 9 = 0 We know that, for x = -3 to be solution of the given quadratic equation it should satisfy the equation. Now, on substituting x = 5 in the given equation, we have L.H.S = 2(-3) 2 – 7(-3) + 9 = 18 + 21 + 9 = 48 ≠ R.H.S Therefore, x = -3 is not a solution of the given quadratic equation 2x 2 – 7x + 9 = 0.1. 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 + 2√3x – 9 = 0
(v) x 2 – ax – b 2 = 0 (vi) 2x 2 + 8x + 9 = 0
Solution:
(i) Given quadratic equation, 7x 2 – 9x + 2 = 0 Here, a = 7, b = -9 and c = 2 So, the Discriminant (D) = b 2 – 4ac D = (-9) 2 – 4(7)(2) = 81 – 56 = 25 As D > 0, the roots of the equation is real and unequal.2. Find the value of ‘p’, if the following quadratic equations has equal roots:
(i) 4x 2 – (p – 2)x + 1 = 0
(ii) x 2 + (p – 3)x + p = 0
Solution:
(i) 4x 2 – (p – 2)x + 1 = 0 Here, a = 4, b = -(p – 2), c = 1 Given that the roots are equal, So, Discriminant = 0 ⇒ b 2 – 4ac = 0 D = (-(p – 2)) 2 – 4(4)(1) = 0 ⇒ p 2 + 4 – 4p – 16 = 0 ⇒ p 2 – 4p – 12 = 0 ⇒ p 2 – 6p + 2p – 12 = 0 ⇒ p(p – 6) + 2(p – 6) = 0 ⇒ (p + 2)(p – 6) = 0 ⇒ p + 2 = 0 or p – 6 = 0 Hence, p = -2 or p = 6Solve equations, number 1 to 20, given below, using factorization method:
1. x 2 – 10x – 24 = 0
Solution:
Given equation, x 2 – 10x – 24 = 0 x 2 – 12x + 2x – 24 = 0 x(x – 12) + 2(x – 12) = 0 (x + 2)(x – 12) = 0 So, x + 2 = 0 or x – 12 = 0 Hence, x = -2 or x = 122. x 2 – 16 = 0
Solution:
Given equation, x 2 – 16 = 0 x 2 + 4x – 4x + 16 = 0 x(x + 4) -4(x + 4) = 0 (x – 4) (x + 4) = 0 So, (x – 4) = 0 or (x + 4) = 0 Hence, x = 4 or x = -43. 2x 2 – ½ x = 0
Solution:
Given equation, 2x 2 – ½ x = 0 4x 2 – x = 0 x(4x – 1) = 0 So, either x = 0 or 4x – 1 = 0 Hence, x = 0 or x = ¼4. x(x – 5) = 24
Solution:
Given equation, x(x – 5) = 24 x 2 – 5x = 24 x 2 – 5x – 24 = 0 x 2 – 8x + 3x – 24 = 0 x(x – 8) + 3(x – 8) = 0 (x + 3)(x – 8) = 0 So, x + 3 = 0 or x – 8 = 0 Hence, x = -3 or x = 85. 9/2 x = 5 + x 2
Solution:
Given equation, 9/2 x = 5 + x 2 On multiplying by 2 both sides, we have 9x = 2(5 + x 2 ) 9x = 10 + 2x 2 2x 2 – 9x + 10 = 0 2x 2 – 4x – 5x + 10 = 0 2x(x – 2) – 5(x – 2) = 0 (2x – 5)(x -2) = 0 So, 2x – 5 = 0 or x – 2 = 0 Hence, x = 5/2 or x = 26. 6/x = 1 + x
Solution:
Given equation, 6/x = 1 + x On multiplying by x both sides, we have 6 = x(1 + x) 6 = x + x 2 x 2 + x – 6 = 0 x 2 + 3x – 2x – 6 = 0 x(x + 3) – 2(x + 3) = 0 (x – 2) (x + 3) = 0 So, x – 2 = 0 or x + 3 = 0 Hence, x = 2 or x = -37. x = (3x + 1)/ 4x
Solution:
Given equation, x = (3x + 1)/ 4x On multiplying by 4x both sides, we have 4x(x) = 3x + 1 4x 2 = 3x + 1 4x 2 – 3x – 1 = 0 4x 2 – 4x + x – 1 = 0 4x(x – 1) + 1(x – 1) = 0 (4x + 1) (x – 1) = 0 So, 4x + 1 = 0 or x – 1 = 0 Hence, x = -1/4 or x = 18. x + 1/x = 2.5
Solution:
Given equation, x + 1/x = 2.5 x + 1/x = 5/2 Taking LCM on L.H.S, we have (x 2 + 1)/ x = 5/2 2(x 2 + 1) = 5x 2x 2 + 2 = 5x 2x 2 – 5x + 2 = 0 2x 2 – 4x – x + 2 = 0 2x(x – 2) -1(x – 2) = 0 (2x – 1)(x – 2) = 0 So, 2x – 1 = 0 or x – 2 = 0 Hence, x = ½ or x = 29. (2x – 3) 2 = 49
Solution:
Given equation, (2x – 3) 2 = 49 Expanding the L.H.S, we have 4x 2 – 12x + 9 = 49 4x 2 – 12x – 40 = 0 Dividing by 4 on both side x 2 – 3x – 10 = 0 x 2 – 5x + 2x – 10 = 0 x(x – 5) + 2(x – 5) = 0 (x + 2) (x – 5) = 0 So, x + 2 = 0 or x – 5 = 0 Hence, x = -2 or 510. 2(x 2 – 6) = 3(x – 4)
Solution:
Given equation, 2(x 2 – 6) = 3(x – 4) 2x 2 – 12 = 3x – 12 2x 2 = 3x x(2x – 3) = 0 So, x = 0 or (2x – 3) = 0 Hence, x = 0 or x = 3/211. (x + 1) (2x + 8) = (x + 7) (x + 3)
Solution:
Given equation, (x + 1) (2x + 8) = (x + 7) (x + 3) 2x 2 + 2x + 8x + 8 = x 2 + 7x + 3x + 21 2x 2 + 10x + 8 = x 2 + 10x + 21 x 2 = 21 – 8 x 2 – 13 = 0 (x – √13) (x + √13) = 0 So, x – √13 = 0 or x + √13 = 0 Hence, x = – √13 or x = √1312. x 2 – (a + b)x + ab = 0
Solution:
Given equation, x 2 – (a + b)x + ab = 0 x 2 – ax – bx + ab = 0 x(x – a) – b(x – a) = 0 (x – b) (x – a) = 0 So, x – b = 0 or x – a = 0 Hence, x = b or x = a13. (x + 3) 2 – 4(x + 3) – 5 = 0
Solution:
Given equation, (x + 3) 2 – 4(x + 3) – 5 = 0 (x 2 + 9 + 6x) – 4x – 12 – 5 = 0 x 2 + 2x – 8 = 0 x 2 + 4x – 2x – 8 = 0 x(x + 4) – 2(x – 4) = 0 (x – 2)(x + 4) = 0 So, x – 2 = 0 or x + 4 = 0 Hence, x = 2 or x = -414. 4(2x – 3) 2 – (2x – 3) – 14 = 0
Solution:
Given equation, 4(2x – 3) 2 – (2x – 3) – 14 = 0 Let substitute 2x – 3 = y Then the equation becomes, 4y 2 – y – 14 = 0 4y 2 – 8y + 7y – 14 = 0 4y(y – 2) + 7(y – 2) = 0 (4y + 7)(y – 2) = 0 So, 4y + 7 = 0 or y – 2 = 0 Hence, y = -7/4 or y = 2 But we have taken y = 2x – 3 Thus, 2x – 3 = -7/4 or 2x – 3 = 2 2x = 5/ 4 or 2x = 5 x = 5/8 or x = 5/215. 3x – 2/ 2x- 3 = 3x – 8/ x + 4
Solution:
Given equation, 3x – 2/ 2x- 3 = 3x – 8/ x + 4 On cross-multiplying we have, (3x – 2)(x + 4) = (3x – 8)(2x – 3) 3x 2 – 2x + 12x – 8 = 6x 2 – 16x – 9x + 24 3x 2 + 10x – 8 = 6x 2 – 25x + 24 3x 2 – 35x + 32 = 0 3x 2 – 3x – 32x + 32 = 0 3x(x – 1) – 32(x – 1) = 0 (3x – 32)(x – 1) = 0 So, 3x – 32 = 0 or x – 1 = 0 Hence, x = 32/3 or x = 116. 2x 2 – 9x + 10 = 0, when:
(i) x ∈ N (ii) x ∈ Q
Solution:
Given equation, 2x 2 – 9x + 10 = 0 2x 2 – 4x – 5x + 10 = 0 2x(x – 2) – 5(x – 2) = 0 (2x – 5)(x – 2) = 0 So, 2x – 5 = 0 or x – 2 = 0 Hence, x = 5/2 or x = 2 (i) When x ∈ N x = 2 is the solution. (ii) When x ∈ Q x = 2, 5/2 are the solutions
1. Solve, each of the following equations, using the formula:
(i) x 2 – 6x = 27
Solution:
Given equation, x 2 – 6x = 27 x 2 – 6x – 27 = 0 Here, a = 1 , b = -6 and c = -27 By quadratic formula, we have
Therefore, x = 9 or -3
(ii) x 2 – 10x + 21 = 0
Solution:
Given equation, x 2 – 10x + 21 = 0 Here, a = 1, b = -10 and c = 21 By quadratic formula, we have
Therefore, x = 7 or x = 3
(iii) x 2 + 6x – 10 = 0
Solution:
Given equation, x 2 + 6x – 10 = 0 Here, a = 1, b = 6 and c = -10 By quadratic formula, we have
Therefore, x = -3 + √19 or x = -3 – √19
(iv) x 2 + 2x – 6 = 0
Solution:
Given equation, x 2 + 2x – 6 = 0 Here, a = 1, b = 2 and c = -6 By quadratic formula, we have
Therefore, x = -1 + √7 or x = -1 – √7
(v) 3x 2 + 2x – 1 = 0
Solution:
Given equation, 3x 2 + 2x – 1 = 0 Here, a = 3, b = 2 and c = -1 By quadratic formula, we have
Therefore, x = 1/3 or x = -1
(vi) 2x 2 + 7x + 5 = 0
Solution:
Given equation, 2x 2 + 7x + 5 = 0 Here, a = 2, b = 7 and c = 5 By quadratic formula, we have
(vii) 2/3 x = -1/6 x 2 – 1/3
Solution:
Given equation, 2/3 x = -1/6 x 2 – 1/3 1/6 x 2 + 2/3 x + 1/3 = 0 Multiplying by 6 on both sides x 2 + 4x + 2 = 0 Here, a = 1, b = 4 and c = 2 By quadratic formula, we have
Therefore, x = -2 + √2 or x = -2 – √2
(viii) 1/15 x 2 + 5/3 = 2/3 x
Solution:
Given equation, 1/15 x 2 + 5/3 = 2/3 x 1/15 x 2 – 2/3 x + 5/3 = 0 Multiplying by 15 on both sides x 2 – 10x + 25 = 0 Here, a = 1, b = -10 and c = 25 By quadratic formula, we have
(ix) x 2 – 6 = 2 √2 x
Solution:
Given equation, x 2 – 6 = 2 √2 x x 2 – 2√2 x – 6 = 0 Here, a = 1, b = -2√2 and c = -6 By quadratic formula, we have
Therefore, x = 3√2 or x = -√2
(x) 4/x – 3 = 5/ (2x + 3)
Solution:
Given equation, 4/x – 3 = 5/ (2x + 3) (4 – 3x)/ x = 5/ (2x + 3) On cross multiplying, we have (4 – 3x)(2x + 3) = 5x 8x – 6x 2 + 12 – 9x = 5x 6x 2 + 6x – 12 = 0 Dividing by 6, we get x 2 + x – 2 = 0 Here, a = 1, b = 1 and c = -2 By quadratic formula, we have
Therefore, x = 1 or x = -2
(xi) 2x + 3/ x + 3 = x + 4/ x + 2
Solution:
Given equation, 2x + 3/ x + 3 = x + 4/ x + 2 On cross-multiplying, we have (2x + 3) (x + 2) = (x + 4) (x + 3) 2x 2 + 4x + 3x + 6 = x 2 + 3x + 4x + 12 2x 2 + 7x + 6 = x 2 + 7x + 12 x 2 + 0x – 6 = 0 Here, a = 1, b = 0 and c = -6 By quadratic formula, we have Therefore, x = √6 or x = -√6(xii) √6x 2 – 4x – 2√6 = 0
Solution:
Given equation, √6x 2 – 4x – 2√6 = 0 Here, a = √6, b = -4 and c = -2√6 By quadratic formula, we have
Therefore, x = √6 or -√6/3
(xiii) 2x/ x – 4 + (2x – 5)/(x – 3) = 
Solution:
Given equation, 2x/ x – 4 + (2x – 5)/(x – 3) =
25x 2 – 175x + 300 = 12x 2 – 57x + 60 13x 2 – 118x + 240 = 0 Here, a = 13, b = -118 and c = 240 By quadratic formula, we have
Therefore, x = 6 or x = 40/13
(xiv)
Solution:
From the given equation,
10x 2 – 60x + 80 = 6x 2 – 30x + 30 4x 2 – 30x + 50 = 0 2x 2 – 15x + 25 = 0 Here, a = 2, b = -15 and c = 25
Therefore, x = 5 or x = 5/2
2. Solve each of the following equations for x and give, in each case, your answer correct to one decimal place:
(i) x 2 – 8x +5 = 0
(ii) 5x 2 + 10x – 3 = 0
Solution:
(i) x 2 – 8x + 5 = 0 Here, a = 1, b = -8 and c = 5 By quadratic formula, we have
x = 4 ± 3.3 Thus, x = 7.7 or x = 0.7 (ii) 5x 2 + 10x – 3 = 0 Here, a = 5, b = 10 and c = -3 By quadratic formula, we have
Thus, x = 0.3 or x = -2.3
3. Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places:
(i) 2x 2 – 10x + 5 = 0
Solution:
Given equation, 2x 2 – 10x + 5 = 0 Here, a = 2, b = -10 and c = 5
Therefore, x = 4.44 or x = 0.56
(ii) 4x + 6/x + 13 = 0
Solution:
Given equation, 4x + 6/x + 13 = 0 Multiplying by x both sides, we get 4x 2 + 13x + 6 = 0 Here, a = 4, b = 13 and c = 6
Therefore, x = -0.56 or x = -2.70
(iii) 4x 2 – 5x – 3 = 0
Solution:
Given equation, 4x 2 – 5x – 3 = 0 Here, a = 4, b = -5 and c = -3
Therefore, x = 1.70 or x = -0.44
(iv) x 2 – 3x – 9 = 0
Solution:
Given equation, x 2 – 3x – 9 = 0 Here, a = 1, b = -3 and c = -9
Therefore, x = 4.85 or x = -1.85
(v) x 2 – 5x – 10 = 0
Solution:
Given equation, x 2 – 5x – 10 = 0 Here, a = 1, b = -5 and c = -10
Therefore, x = 6.53 or x = -1.53
4. Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places:
(i) 3x 2 – 12x – 1 = 0
(ii) x 2 – 16 x +6 = 0
(iii) 2x 2 + 11x + 4 = 0
Solution:
(i) Given equation, 3x 2 – 12x – 1 = 0 Here, a = 3, b = -12 and c = -1
Therefore, x = 15.616 or x = 0.384 (iii) Given equation, 2x 2 + 11x + 4 = 0 Here, a = 2, b = 11 and c = 4
Therefore, x = -0.392 or x = -5.110
5. Solve:
(i) x 4 – 2x 2 – 3 = 0
Solution:
Given equation, x 4 – 2x 2 – 3 = 0 x 4 – 3x 2 + x 2 – 3 = 0 x 2 (x 2 – 3) + 1(x 2 – 3) = 0 (x 2 + 1) (x 2 – 3) = 0 So, x 2 + 1 = 0 (which is not possible) or x 2 – 3 = 0 Hence, x 2 – 3 = 0 x = ± √3(ii) x 4 – 10x 2 + 9 = 0
Solution:
Given equation, x 4 – 10x 2 + 9 = 0 x 4 – x 2 – 9x 2 + 9 = 0 x 2 (x 2 – 1) – 9(x 2 – 1) = 0 (x 2 – 9)(x 2 – 1) = 0 So, we have x 2 – 9 = 0 or x 2 – 1 = 0 Hence, x = ± 3 or x = ± 11. Solve each of the following equations:
Solution:
Given equation,
4x 2 + 6x + x – 3 + 3x + 9 = 0 4x 2 + 10x + 6 = 0 4x 2 + 4x + 6x + 6 = 0 4x(x + 1) + 6(x + 1) = 0 (4x + 6) (x + 1) = 0 So, 4x + 6 = 0 or x + 1 = 0 x = -1 or x = -6/4 = -3/2 (rejected as this value is excluded in the domain) Therefore, x = -1 is the only solution
2. (2x + 3) 2 = 81
Solution:
Given, (2x + 3) 2 = 81 Taking square root on both sides we have, 2x + 3 = ± 9 2x = ± 9 – 3 x = (± 9 – 3)/ 2 So, x = (9 – 3)/ 2 or (-9 – 3)/ 2 Therefore, x = 3 or x = -63. a 2 x 2 – b 2 = 0
Solution:
Given equation, a 2 x 2 – b 2 = 0 (ax) 2 – b 2 = 0 (ax + b)(ax – b) = 0 So, ax + b = 0 or ax – b = 0 Therefore, x = -b/a or b/a4. x 2 – 11/4 x + 15/8 = 0
Solution:
Given equation, x 2 – 11/4 x + 15/8 = 0 Taking L.C.M we have, (8x 2 – 22x + 15)/ 8 = 0 8x 2 – 22x + 15 = 0 8x 2 – 12x – 10x + 15 = 0 4x(2x – 3) – 5(2x – 3) = 0 (4x – 5)(2x – 3) = 0 So, 4x – 5 = 0 or 2x – 3 = 0 Therefore, x = 5/4 or x = 3/25. x + 4/x = -4; x ≠ 0
Solution:
Given equation, x + 4/x = -4 (x 2 + 4)/ x = -4 x 2 + 4 = -4x x 2 + 4x + 4 = 0 x 2 + 2x + 2x + 4 = 0 x(x + 2) + 2(x + 2) = 0 (x + 2)(x + 2) = 0 (x + 2) 2 = 0 Taking square – root we have, x + 2 = 0 Therefore, x = -26. 2x 4 – 5x 2 + 3 = 0
Solution:
Given equation, 2x 4 – 5x 2 + 3 = 0 Let’s take x 2 = y Then, the equation becomes 2y 2 – 5y + 3 = 0 2y 2 – 2y – 3y + 3 = 0 2y(y – 1) – 3(y – 1) = 0 (2y – 3) (y – 1) = 0 So, 2y – 3 = 0 or y – 1 = 0 y = 3/2 or y = 1 And, we have taken y = x 2 Thus, x 2 = 3/2 or x 2 = 1 x = ± √(3/2) or x = ±17. x 4 – 2x 2 – 3 = 0
Solution:
Given equation, x 4 – 2x 2 – 3 = 0 Let’s take x 2 = y Then, the equation becomes y 2 – 2y – 3 = 0 y 2 – 3y + y – 3 = 0 y(y – 3) + 1(y – 3) = 0 (y + 1)(y – 3) = 0 So, y + 1 = 0 or y – 3 = 0 y = -1 or y = 3 And, we have taken y = x 2 Thus, x 2 = – 1(impossible, no real solution) x 2 = 3 x = ± √3 8.
Solution:
Let us take (x + 1/x) = y …. (1) Now, squaring it on both sides (x + 1/x) 2 = y 2 x 2 + 1/x 2 + 2 = y 2 So, x 2 + 1/x 2 = y 2 – 2 ….. (2) Using (1) and (2) in the given equation, we have 9(y 2 – 2) – 9(y) – 52 = 0 9y 2 – 18 – 9y – 52 = 0 9y 2 – 9y – 70 = 0 9y 2 – 30y + 21y – 70 = 0 3y(3y – 10) + 7(3y – 10) = 0 (3y + 7)(3y – 10) = 0 So, 3y + 7 = 0 or 3y – 10 = 0 y = -7/3 or y = 10/3 Now, x + 1/x = -7/3 or x + 1/x = 10/3 (x 2 + 1)/x = -7/3 or (x 2 + 1)/x = 10/3 3x 2 – 10x + 3 = 0 or 3x 2 + 7x + 3 = 0
3x 2 – 9x – x + 3 = 0 or 3x(x – 3) – 1(x – 3) = 0 (3x – 1)(x – 3) = 0 So, x = 1/3 or 3
9.
Solution:
Let us take (x + 1/x) = y …. (1) Now, squaring it on both sides (x + 1/x) 2 = y 2 x 2 + 1/x 2 + 2 = y 2 So, x 2 + 1/x 2 = y 2 – 2 ….. (2) Using (1) and (2) in the given equation, we have 2(y 2 – 2) – (y) = 11 2y 2 – 4 – y = 11 2y 2 – y – 15 = 0 2y 2 – 6y + 5y – 15 = 0 2y(y – 3) + 5(y – 3) = 0 (2y + 5) (y – 3) = 0 So, 2y + 5 = 0 or y – 3 = 0 y = -5/2 or y = 3 Now, x + 1/x = -5/2 or x + 1/x = 3 (x 2 + 1)/x = -5/2 or (x 2 + 1)/x = 3 2(x 2 + 1) = -5x or x 2 + 1 = 3x 2x 2 + 5x + 2 = 0 or x 2 – 3x + 1 = 0
2x 2 + 4x + x + 2 = 0 or 2x(x + 2) + 1(x + 2) = 0 (2x + 1)(x + 2) = 0 Hence, x = -1/2 or -2
10.
Solution:
Let us take (x – 1/x) = y …. (1) Now, squaring it on both sides (x – 1/x) 2 = y 2 x 2 + 1/x 2 – 2 = y 2 So, x 2 + 1/x 2 = y 2 + 2 ….. (2) Using (1) and (2) in the given equation, we have (y 2 + 2) – 3(y) – 2 = 0 y 2 -3y = 0 y(y – 3) = 0 So, y = 0 or y – 3 = 0 Now, (x – 1/x) = 0 or (x – 1/x) = 3 x 2 – 1 = 0 or x 2 – 1 = 3x x 2 = 1 or x 2 – 3x – 1 = 0
x = ± 1 or
1. Solve:
(i) (x + 5) (x – 5) = 24
Solution:
Given equation, (x + 5) (x – 5) = 24 x 2 – 25 = 24 x 2 = 49 Thus, x = ± 7(ii) 3x 2 – 2√6x + 2 = 0
Solution:
Given equation, 3x 2 – 2√6x + 2 = 0 3x 2 – √6x – √6x + 2 = 0 √3x(√3x – √2) – √2(√3x – √2) = 0 (√3x – √2) (√3x – √2) = 0 So, √3x – √2 = 0 or √3x – √2 = 0 Therefore, x = √(2/3), √(2/3) (equal roots)(iii) 3√2x 2 – 5x – √2 = 0
Solution:
Given equation, 3√2x 2 – 5x – √2 = 0 3√2x 2 – 6x + x – √2 = 0 3√2x(x – √2) + 1(x – √2) = 0 (3√2x + 1) (x – √2) = 0 So, 3√2x + 1 = 0 or x – √2 = 0 Therefore, x = -1/ 3√2 or x = √2(iv) 2x – 3 = √(2x 2 – 2x + 21)
Solution:
Given equation, 2x – 3 = √(2x 2 – 2x + 21) On squaring on both sides, we have (2x – 3) 2 = 2x 2 – 2x + 21 4x 2 + 9 – 12x = 2x 2 – 2x + 21 2x 2 – 10x – 12 = 0 Dividing by 2, we get x 2 – 5x – 6 = 0 x 2 – 6x + x – 6 = 0 x(x – 6) + 1(x – 6) = 0 (x + 1) (x – 6) = 0 So, x + 1 = 0 or x – 6 = 0 Thus, we get x = -1 or x = 6 But, putting x = -1 the L.H.S become negative. And we know that the square root function always gives a positive value. Therefore, x = 6 is the only solution.2. One root of the quadratic equation 8x 2 + mx + 15 = 0 is ¾. Find the value of m. Also, find the other root of the equation.
Solution:
Given equation, 8x 2 + mx + 15 = 0 One of the roots is ¾, and hence it satisfies the given equation So, 8(3/4) 2 + m(3/4) + 15 = 0 8(9/16) + m(3/4) + 15 = 0 18/4 + 3m/4 + 15 = 0 Taking L.C.M, we have (18 + 3m + 60)/4 = 0 18 + 3m + 60 = 0 3m = – 78 m = -26 Now, putting the value of m in the given equation, we get 8x 2 + (-26)x + 15 = 0 8x 2 – 26x + 15 = 0 8x 2 – 20x – 6x + 15 = 0 4x(2x – 5) – 3(2x – 5) = 0 (4x – 3) (2x – 5) = 0 So, 4x – 3 = 0 or 2x – 5 = 0 Therefore. x = ¾ or x = 5/23. Show that one root of the quadratic equation x 2 + (3 – 2a)x – 6a = 0 is -3. Hence, find its other root.
Solution:
Given quadratic equation, x 2 + (3 – 2a)x – 6a = 0 Now, putting x = -3 we have (-3) 2 + (3 – 2a)( -3) – 6a = 0 9 – 9 + 6a – 6a = 0 0 = 0 Since, x = -3 satisfies the given equation -3 is one of the root of the quadratic equation. x 2 + (3 – 2a)x – 6a = 0 x 2 + 3x – 2ax – 6a = 0 x(x + 3) – 2a(x + 3) = 0 (x – 2a) (x + 3) = 0 So, x – 2a = 0 or x + 3 =0 x = 2a or x = -3 Hence, the other root is 2a.4. If p – 15 = 0 and 2x 2 + px + 25 = 0: find the values of x.
Solution:
Given equations, p – 15 = 0 and 2x 2 + px + 25 = 0 Thus, p = 15 Now, using p in the quadratic equation, we get 2x 2 + (15)x + 25 = 0 2x 2 + 10x + 5x + 25 = 0 2x(x + 5) + 5(x + 5) = 0 (2x + 5) (x + 5) = 0 So, 2x + 5 = 0 or x + 5 = 0 Hence, x = -5/2 or x = -55. Find the solution of the quadratic equation 2x 2 – mx – 25n = 0; if m + 5 = 0 and n – 1 = 0.
Solution:
Given, m + 5 = 0 and n – 1 = 0 so, m = -5 and n = 1 Now, putting these values in the given quadratic equation 2x 2 – mx – 25n = 0, we get 2x 2 – (-5)x – 25(1) = 0 2x 2 + 5x – 25 = 0 2x 2 + 10x – 5x – 25 = 0 2x(x + 5) -5(x + 5) = 0 (2x – 5) (x + 5) = 0 So, 2x – 5 = 0 or x + 5 = 0 Hence, x = 5/2 or x = -56. If m and n are roots of the equation: 1/x – 1/(x-2) = 3: where x ≠ 0 and x ≠ 2; find m x n.
Solution:
Given equation, 1/x – 1/(x-2) = 3 (x – 2 – x)/ (x(x – 2)) = 3 -2 = 3(x 2 – 2x) 3x 2 – 6x + 2 = 0 Solving by using quadratic formula, we get
And, since m and n are roots of the equation, we have m = (√3 + 1)/ √3 n = (√3 – 1)/ √3 So, m x n = (√3 + 1)/ √3 x (√3 – 1)/ √3 = [(√3) 2 – 1]/ (√3) 2 Thus, m x n = 2/3
With the ICSE Class 10 board exams having started from 17 February 2026, smart and focused revision is the key to scoring well. Quadratic Equations is an important and high-weightage chapter, so clarity in methods and formulas is essential.
Revise all three methods – factorisation, completing the square, and quadratic formula.
Memorise the quadratic formula correctly and practice substituting values carefully.
Check the discriminant (D = b² – 4ac) to identify the nature of roots.
Practice 3–4 important sums to improve speed and accuracy.
Avoid calculation mistakes by solving step-by-step and rechecking signs.
