Question

If one zero of p(x) = 4x2 – (8k2 – 40k) x – 9 is negative of the other, find values of k.

Answer 1

Given :-

p ( x ) = 4x² - ( 8k² - 40k )x - 9

Let the zeros be a and -a

then ,

Sum of zeros =

$\frac{ - coeff \: of \: x}{coeff \: of \: {x}^{2} }$

$a \: + ( - a) = \frac{ (8 {k}^{2} - 40k) }{4}$

$0 = \frac{8 {k}^{2} - 40k }{4}$

0 = 8k² - 40k

0 = k² - 5k

0 = k ( k - 5 )

k = 0

k = 5



@Altaf

Answer 2

hope this helps you....any doubt u can ask.....