Question

Find the value of k for which the roots of the quadratic equation (k-4)x^2+2(k-4)x+2=0 are equal

Answer 1

SOLUTION ☺️

$= > (k - 4)x {}^{2} + 2(k - 4)x +2 = 0 \\ = > a = k - 4 \\ = > b = 2(k - 4) \\ = > c = 2 \\ = > since \: roots \: are \: equal \\ = > discriminate = 0 \\ = > b {}^{2} - 4ac = 0 \\ = > 2(k - 4) {}^{2} - 4(k - 4)(2) = 0 \\ = > 4(k {}^{2} - 8k + 16) - 8(k - 4) = 0 \\ = > 4k {}^{2} - 32k + 64 - 8k + 32 = 0 \\ = > 4k {}^{2} - 40k + 96 = 0 \\ = > k {}^{2} - 10k + 24 = 0 \\ = >k {}^{2} - 6k - 4k + 24 = 0 \\ = > k(k - 6) - 4(k - 6) = 0 \\ = > (k - 6) = 0 \: \: or(k - 4) = 0 \\ = > k = 6 \: \: or \: k = 4$

HOPE it helps ✔️