ICSE Class 10 Maths Selina Solutions Chapter 5: ICSE Class 10 Maths Selina Solutions for Chapter 5, "Quadratic Equations," make it easy to understand and solve quadratic equations.
It covers different methods to find the solutions, like factoring, completing the square, and using the quadratic formula. The solutions are explained step-by-step, helping students see how to apply these methods and solve problems correctly. By using these solutions students can get better at solving quadratic equations and be well-prepared for their exams.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. (ii) 5x 2 – 8x = -3(7 – 2x) ⇒ 5x 2 – 8x = 6x – 21 ⇒ 5x 2 – 14x + 21 = 0; which is of the general form ax 2 + bx + c = 0. Thus, the given equation is a quadratic equation. (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 general form ax 2 + bx + c = 0. And it’s a linear equation. Thus, the given equation is not a quadratic equation. (iv) x 2 + 5x – 5 = (x – 3) 2 ⇒ x 2 + 5x – 5 = x 2 – 6x + 9 ⇒ 11x – 14 = 0; which is not of the general form ax 2 + bx + c = 0. And it’s a linear equation. Thus, the given equation is not a quadratic equation. (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 general form ax 2 + bx + c = 0. And it’s a cubic equation. Thus, the given equation is not a quadratic equation. (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 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. (ii) Given quadratic equation, 6x 2 – 13x + 4 = 0 Here, a = 6, b = -13 and c = 4 So, the Discriminant (D) = b 2 – 4ac D = (-13) 2 – 4(6)(4) = 169 – 96 = 73 As D > 0, the roots of the equation is real and unequal. (iii) Given quadratic equation, 25x 2 – 10x + 1 = 0 Here, a = 25, b = -10 and c = 1 So, the Discriminant (D) = b 2 – 4ac D = (-10) 2 – 4(25)(1) = 100 – 100 = 0 As D = 0, the roots of the equation is real and equal. (iv) Given quadratic equation, x 2 + 2√3x – 9 = 0 Here, a = 1, b = 2√3 and c = -9 So, the Discriminant (D) = b 2 – 4ac D = (2√3) 2 – 4(1)(-9) = 12 + 36 = 48 As D > 0, the roots of the equation is real and unequal. (v) Given quadratic equation, x 2 – ax – b 2 = 0 Here, a = 1, b = -a and c = -b 2 So, the Discriminant (D) = b 2 – 4ac D = (a) 2 – 4(1)(-b 2 ) = a 2 + 4b 2 a 2 + 4b 2 is always positive value. Thus D > 0, and the roots of the equation is real and unequal (vi) Given quadratic equation, 2x 2 + 8x + 9 = 0 Here, a = 2, b = 8 and c = 9 So, the Discriminant (D) = b 2 – 4ac D = (8) 2 – 4(2)(9) = 64 – 72 = -8 As D < 0, the equation has no roots.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 = 6 (ii) x 2 + (p – 3)x + p = 0 Here, a = 1, b = (p – 3), c = p Given that the roots are equal, So, Discriminant = 0 ⇒ b 2 – 4ac = 0 D = (p – 3) 2 – 4(1)(p) = 0 ⇒ p 2 + 9 – 6p – 4p = 0 ⇒ p 2 – 10p + 9 = 0 ⇒ p 2 -9p – p + 9 = 0 ⇒ p(p – 9) – 1(p – 9) = 0 ⇒ (p -9)(p – 1) = 0 ⇒ p – 9 = 0 or p – 1 = 0 Hence, p = 9 or p = 1Solve 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 solutions1. 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(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(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(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(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(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(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(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(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
(xiii) 2x/ x – 4 + (2x – 5)/(x – 3) =
Solution:
Given equation, 2x/ x – 4 + (2x – 5)/(x – 3) =
(xiv)
Solution:
From the given equation,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 have3. 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(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(iii) 4x 2 – 5x – 3 = 0
Solution:
Given equation, 4x 2 – 5x – 3 = 0 Here, a = 4, b = -5 and c = -3(iv) x 2 – 3x – 9 = 0
Solution:
Given equation, x 2 – 3x – 9 = 0 Here, a = 1, b = -3 and c = -9(v) x 2 – 5x – 10 = 0
Solution:
Given equation, x 2 – 5x – 10 = 0 Here, a = 1, b = -5 and c = -104. 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
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,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
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
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 = 01. 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