The value of k for which the equation (K - 2) | vedclass

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(B) For the quadratic equation $(K - 2)x^2 + 8x + K + 4 = 0$ to have real and distinct roots,the discriminant $D$ must be greater than $0$.$D = b^2 -

Vedclass · Accepted Answer

$3$

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(B) For the quadratic equation $(K - 2)x^2 + 8x + K + 4 = 0$ to have real and distinct roots,the discriminant $D$ must be greater than $0$.$D = b^2 - 4ac = 8^2 - 4(K - 2)(K + 4) > 0$$64 - 4(K^2 + 2K - 8) > 0$$16 - (K^2 + 2K - 8) > 0$$-K^2 - 2K + 24 > 0 \Rightarrow K^2 + 2K - 24 $(K + 6)(K - 4) For the roots to be negative,the product of roots $\alpha\beta = \frac{c}{a} = \frac{K + 4}{K - 2} > 0$ and the sum of roots $\alpha + \beta = -\frac{b}{a} = -\frac{8}{K - 2} From $\frac{K + 4}{K - 2} > 0$,we get $K  2$.From $-\frac{8}{K - 2}  0$,which implies $K > 2$.Combining the conditions $-6  2$,we get $2 Among the given options,only $K = 3$ satisfies this condition.

The value of $k$ for which the equation $(K - 2)x^2 + 8x + K + 4 = 0$ has both roots real,distinct and negative is

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