Question

If the sum of the squares of the roots of the equation x^2+2x-p=0 is 8, then find the value of p

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Answer 1

Answer:

P(x)=x²+2x-p

This is in the form ax²+bx+c.

So a=1,b=2,c=-p

Now, let the zeroes be a and b (u can take alpha and beta).

Given, a²+b²=8

(a+b)² - 2ab = 8

Now , Sum of roots = a+b= -b/a = -2

Product of roots = ab = c/a = -p

So, by putting the values we get,

(-2)² - 2(-p) = 8

4+2p=8

2p=8–4

p=4/2

p=2.

Answer 2

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