Question

For what value of k , (4-k)x²+ (2k+4)x + (8k+1) =0 is a perfect square.

Answer 1

$\large{\mathbf{HELLO \: \: DEAR,}}$

$\mathbf{GIVEN \: \: THAT:- }\\ \\ \mathbf{b^2 - 4ac = 0 = D^2} \\ \\ (4-k) {x}^{2} + (2k+4)x + (8k+1) =0 \\ \\ \mathbf{WHERE ,} \\ \\ \mathbf{a = (4 - k) , b = (2k + 4) , c = (8k + 1)}\\ \\ \mathbf{(2k + 4)^2 - 4[(4 - k)*(8k + 1)] = 0} \\ \\ \mathit{4k^2 + 16 + 16k - 4[32k + 4 - 8k^2 - k] = 0}\\ \\ \mathit{4k^2 + 16 + 16k - 128k - 16 + 32k^2 + 4k = 0}\\ \\ \mathit{32k^2 - 108 = 0}\\ \\ \mathit{9k^2 - 27k = 0}\\ \\ \mathit{k = 3}$

$\large{\mathbf{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$

Answer 2

Hope it helps .....................