InterviewSolution
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Find value of k for which following quadratic equations have real and fractional roots.(i) kx2 + 2x + 1 = 0(ii) kx2 + 6x + 1 = 0(iii) x2 – kx + 9 = 0 |
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Answer» (i) Given roots kx2 + 2x + 1 = 0 Comparing it by general quadratic equation ax2 + bx + c = 0 a = k, b = 2, c = 1 Discriminant (D) = b2 – 4ac = (2)2 – 4 × k × 1 = 4 – 4 k = 4(1 – k) For real and fractional roots Discriminant (D) > 0 ⇒ 4(1 -k) > 0 ⇒ 1 – k > 0 ⇒ k < 1 Thus, k will be less than. (ii) Given equation kx2 + 6x + 1 = 0 Comparing it by general quadratic equation ax2 + bx + c – 0, a = k, b = 6, c = 1 Discriminant (D) = b2 – 4ac = (6)2 – 4 × k × 1 = 36 – 4k = 4(9 – k) If discriminant (D) > 0 then roots of equation will be real and fractional. i.e., D > 0 ⇒ 4(9 – k) > 0 ⇒ 9 – k > 0 ⇒ k > 9 Thus value of k will be greater than 9 (iii) Given equation x2 – kx + 9 = 0 Comparing it by general quadratic equation ax2 + bx + c = 0 a = 1, b = -k, c = 9 Discriminant (D) = b2 – 4ac = (-k)2 – 4 × 1 × 9 = k2 – 36 If D > 0, then roots of equation will be real and fractional. D > 0 ⇒ k2 – 36 > 0 ⇒ (k – 6)(k + 6) > 0 ⇒ k – 6 > 0 or k + 6 > 0 ⇒ k > 6 or k < -6 Thus, value of k will be greater than 6 or less than -6 |
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