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For what value of k+4x2 + k+1x + 1 has equal roots

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STEP by step SOLUTION:

Given \ p(x) \ (k+4)x^{2}+(k+1)x+1 \ has \ a\ equal \ roots.\\\\For \ equal \ roots=b^{2}-4ac=0\\\\putting \ value \ here\\\\(k+1)^{2}-4(k+4) \times1=0\\\\k^{2}+ 1^{2}+2k=4k+16\\\\k^{2}-2k-15=0\\\\Now \ factorize \ it \ by \ formula\\\\x=\frac{-b+-\sqrt{b^{2}-4ac} }{2a}\\\\x=\frac{2+-8}{2}\\\\x=\frac{10}{2}=5 \ or \ x=\frac{-6}{2}=-3

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