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Clear selection2 points38. If x2 + k(4x + k – 1) + 2 = 0 has equal roots, then k =(a)(b) ,-1(c)3 1(d)ilm12 3аObOCOdClear selection |
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Answer» SOLUTION : Option (b) is correct : ⅔ , - 1 GIVEN : x² + k(4x + k - 1) + 2 = 0 x² + 4kx + k² - k + 2 = 0 On comparing the given EQUATION with ax² + BX + c = 0 Here, a = 1 , b = 4k , c = k² - k + 2 D(discriminant) = b² – 4ac D = (4k)² - 4 × 1 × (k² - k + 2) D = 16k² - 4k² + 4k - 8 D = 12k² + 4k - 8 D = 4(3k² + k - 2) D = 0 (equal roots given) 4(3k² + k - 2) = 0 (3k² + k - 2) = 0 3k² + 3k - 2k - 2 = 0 [By middle term splitting] 3k (k + 1) - 2(k + 1) = 0 (3k - 2)(k + 1) = 0 (3k - 2) = 0 (k + 1) = 0 3k = 2 or k = - 1 k = ⅔ or k = - 1 Hence, the value of k is ⅔ & - 1. HOPE THIS ANSWER WILL HELP YOU. |
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