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

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(C) For a quadratic equation $ax^2 + bx + c = 0$ to have real and distinct roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$.Here,$a = (

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$3$

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(C) For a quadratic equation $ax^2 + bx + c = 0$ to have real and distinct roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$.Here,$a = (k - 2)$,$b = 8$,and $c = (k + 4)$.$D = 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 roots to be negative,the sum of roots $-b/a = -8/(k - 2) 0$.From $-8/(k - 2) 0 \Rightarrow k > 2$.From $(k + 4)/(k - 2) > 0$,we get $k > 2$ or $k Combining these conditions with $-6 Checking the options,$k = 3$ satisfies $2 < 3 < 4$.

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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