Question

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

Answer 1

Hey!


Given Equation :- x² - 8x + ( k + 1 )

And it is given that ,one zero of the equation is double of the other.


Let the 1st zeros be y and second Zeros be 2y


• Sum of the Zeros =
$= \frac{ - coefficient \: \: \: of \: x}{coeff. \: \: \: of \: {x}^{2} }$


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

3y = 8

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

• Product of Zeros =
$= \frac{constant \: \: term}{coeff. \: \: of \: {x}^{2} }$


$y \times 2y = \frac{k + 1}{1}$

$2 {y}^{2} = \frac{k + 1}{1}$


$2 \times \frac{8}{3} \times \frac{8}{3} = k + 1$


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


128 = 9k + 9

128 - 9 = 9k

119 = 9k

119/9 = k !!

Answer 2

Here is your Answer

I hope this will help you