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Find the value of k for which the given quadratic equation x2-4x+k=0 has distinct real roots. |
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Answer» For distinct real roots the discriminant should be >0 (-4)^2-4×1×k>0 => 4k < 16 => k<4 x2 - 4x + k = 0 Here, a= 1, b = -4 and c = k Since, D = b2 - 4ac therefore, (-4)2 - 4(1)(k) > 0 [ Since , there are two distinct roots] 16 - 4k > 0 16 > 4k 16/4 > k k < 4 Hence, when k < 4, then the quadratic equation x2 – 4x + k = 0 has distinct real roots. |
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