

InterviewSolution
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Find the values of k for which the roots are real and equal in each of the following equations:(i) (k + 1) x2 + 2 (k + 3) x + (k + 8) = 0(ii) x2 - 2kx + 7x + 1/4 = 0(iii) (k + 1) x2 - 2 (3k + 1) x + 8k + 1 = 0(iv) 5x2 - 4x + 2 +k ((4x2 - 2x - 1) = 0(v) (4 - k) x2 + (2k + 4) x + (8k + 1) = 0 |
Answer» (i) (k + 1) x2 + 2 (k + 3) x + (k + 8) = 0 For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D = 0, roots are real and equal (k + 1) x2 + 2 (k + 3) x + (k + 8) = 0 ⇒ D = 4(k + 3)2 – 4(k + 1)(k + 8) = 0 ⇒ 4k2 + 36 + 24k – 4k2 – 32 – 36k = 0 ⇒ 12k = 4 ⇒ k = 1/3 (ii) x2 - 2kx + 7x + 1/4 = 0 For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D = 0, roots are real and equal x2 – 2kx + 7x + 1/4 = 0 ⇒ D = (7 – 2k)2 – 4 × 1/4 = 0 ⇒ 49 + 4k2 – 28k – 1 = 0 ⇒ k2 – 7k + 12 = 0 ⇒ k2 – 4k – 3k + 12 = 0 ⇒ k(k – 4) – 3(k – 4) = 0 ⇒ (k – 3)(k – 4) = 0 ⇒ k = 3, 4 (iii) (k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0 For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D = 0, roots are real and equal (k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0 ⇒ D = 4(3k + 1)2 – 4(k + 1)(8k + 1) = 0 ⇒ 4 × (9k2 + 6k + 1) – 32k2 – 4 – 36k = 0 ⇒ 36k2 + 24k + 4 – 32k2 – 4 – 36k = 0 ⇒ 4k(k – 3) = 0 ⇒ k = 0, 3 (iv) 5x2 - 4x + 2 + k (4x2 - 2x - 1) = 0 For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D = 0, roots are real and equal 5x2 - 4x + 2 + k (4x2 - 2x - 1) = 0 ⇒ (5 + 4k)x2 – (4 + 2k)x + 2 – k = 0 ⇒ D = (4 + 2k)2 – 4 × (5 + 4k)(2 – k) = 0 ⇒ 16 + 4k2 + 16k + 16k2 – 12k – 40 = 0 ⇒ 20k2 – 4k – 24 = 0 ⇒ 5k2 - k - 6 = 0 ⇒ 5k2 – 6k + 5k – 6 = 0 ⇒ k(5k – 6) + (5k – 6) = 0 ⇒ (k + 1)(5k – 6) = 0 ⇒ k = -1, 6/5 (v) (4 - k)x2 + (2k + 4)x + (8k + 1) = 0 For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D = 0, roots are real and equal (4 - k)x2 + (2k + 4)x + (8k + 1) = 0 ⇒ D = (2k + 4)2 – 4 × (4 – k)(8k + 1) = 0 ⇒ 4k2 + 16 + 16k + 32k2 – 16 – 124k = 0 ⇒ 36k2 – 108k = 0 ⇒ 36k(k – 3) = 0 ⇒ k = 0, 3 |
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