Question

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

Answer 1

x²-8x+(k+1) =0

let roots are "a&b"

if a = 2b

then sum of roots
= a+b = 3b
&
3b = 8

so
b = 8/3
and
a = 16/3


product of roots
= k+1

ab = k+1
128/9 = k+1
____________
k = 128/9 -1
= 119/9


hope it helps you
@di

Answer 2

given equation:

x²-8x+(k+1)

sum of zeroes:

y+2y=$\frac{8}{1}$

3y=8

y=$\frac{8}{3}$

product of zeroes:

y × 2y = $\frac{k+1}{1}$

2y²=$\frac{k+1}{1}$

2×$\frac{8}{3}$×$\frac{8}{3}$= k+1

$\frac{128}{9}$= k+1

128= 9k+9

128-9=9k

119=9k

k=$\frac{119}{9}$

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