Question

if one zero of p(x) =4x^2- (8k^2- 40k) x-9 is negative of the Other, find values of K.

Answer 1

p(x)=$4 x^{2}-(8k^2-40k)-9$
Let α, β be zeroes of p(x)
ATQ, α=-β
we know Sum of zeroes = $\frac{-b}{a}$
$\alpha + \beta = \frac{(8k^2-40k)}{4}$
[tex] - \beta + \beta= \frac{(8k^2-40k)}{4} [/tex]
$0 = 8k^2-40k$
$k^2 - 5k = 0$
$k(k-5)=0$
$\implies k= 0, 5$