If the sum of the squares of the squares of the roots of `x^(2)-(p-2) x-(p+1) = 0 (p in R) ` is 5,
then what is the value of p?

Options

A. 0
B. -1
C. 1
D. `3/2`

Answer

C

Explanation

Let `alpha and beta " be the roots of " `
`x^(2) - (p-2) x- (p+1) = 0, `
Then ` alpha + beta - (p-2) and alpha beta = - (p+1)`
`:' alpha^(2) + beta^(2) = ( alpha + beta)^(2) - 2 alphabeta`
`rArr ( alpha + beta)^(2) = ( alpha + beta)^(2) - 2alphabeta= 5 `
` rArr (p-2)^(2) + 2 ( p+1) = 5 `
` rArr p^(2) - 4p + 4 + 2p + = 5 `
`rArr p^(2) - 4p + 4 + 2 p + 2 = 5 `
`rArr P^(2) - 2p + 1 = 0 `
`rArr(p-1)^(2) = 0 `
`rArr p= 1`