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

let one root be y and other be -y then sum of roots = -b/a here , 0= 8k^2-40k or 40k=8k^2 or k= 5

Answer 2

Answer:

$Value \:of \:k = 0 \:Or \:k=5$

Step-by-step explanation:

Given one zero of p(x)=4x²-(8k²-40k)x-9 is negative of the other .

$Let \:\alpha \:and -\alpha \\are \:two \: zeroes \:of \:p(x)$

$Compare \:p(x)\:with \\ax^{2}+bx+c, \:we \:get$

$a=4, \:b =-(8k^{2}-40k),\:c=-9$

$Sum \:of\:the\: zeroes =\frac{-b}{a}$

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

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

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

$\implies 8k(k-5)=0$

$\implies 8k=0\:Or \:k-5=0$

$\implies k = 0 \:Or \:k = 5$

Therefore,

$Value \:of \:k = 0 \:Or \:k=5$

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