Question

for what value of k,one zero of x^2-8x+(k+1) is double the other

Answer 1

let roots be
$\alpha \: and \: \: 2 \alpha \\ \\ \alpha + 2 \alpha = 8 \\ \alpha = \frac{8}{3} \\ \alpha \times 2 \alpha = k + 1 \\ 2 (\frac{64}{9} ) = k + 1 \\ k = \frac{128 - 9}{9}$


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Answer 2

$128 - 9 \frac{9}{?}$
ans