Question

For What value of 'k' x2 + k will be a complete square​

Answer 1

Correct Question:

For what value of k,(4−k)x2+2(k+2)x+(8k+1) is a perfect square i.e. has equal roots.

Answer:

$\hookrightarrow$perfect square = equal roots = Δ = 0

∴(2(k+2))2−4(4−k)(8k+1)=0

$:\implies\sf{4(k2+4+4k)-4(32k+4-8k2-k)=0}$

$:\implies\sf{k2+4+4k-32k-4+8k2+k=0}$

$:\implies\sf{9k2-27k=0}$

$:\implies\sf{k2-3k=0}$

$:\implies\sf{k(k-3)=0}$

Therefore,the value is 0 :)