Question

If one zeroes of p(x)=4x^2 -(8k^2-40k)x-9 is negative of the other, find values of K.

Answer 1

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Answer 2

p(x)=$4 x^{2} -(8 k^{2} -40k)x-9$
Let α and β be the zeroes of p(x)
α+β=$\frac{-b}{a}$= $\frac{-(8k^{2} -40k)}{4}$= $\frac{-8k^{2} +40k}{4}$
$\alpha + \beta =-2 k^{2} +10k$
According to the question,
α=-β
∴α+β=0

$-2 k^{2} +10k=0$
$k(-2 k+10)=0$
$-2 k+10=0$
$-2k =-10$
$k= \frac{-10}{-2} = 5$
∴k=5