Question

For what value of p, will the sum of the squares of the roots of x^2-px-1+p=0 be minimum?​

Answer 1

Answer:

Step-by-step explanation:

let m and be the roots of the given quadratic

given in the question is sum (m)^2+(n)^2 to be minimum

but we'll transform m^2+n^2 a bit and write it as

(m^2+n^2+2mn)-2mn

(m+n)^2-mn-----(1)  so from here,

sum of roots will be -b/a but in the given quadratic m+n=p

product of roots will be c/a and in the given quadratic it will be p-1

so plugging this in (1) we get

(p)^2-(p-1) which equals p^2-p+1 so for the minimum value of this we need to differentiate what we got wrt p

so differentiating one time we get 2p-1=0 so p must be equal to 1/2

just to check if this gives the minimum value we need double derivative of the p^2-p+1 to be greater than 0

so double derivative is  2 which is greater than 0 so the p value we need is "1/2"

HOPE THIS HELPS YOU