Question

Determine the value of p if one root of equation x^2+px+1=0 is the square of other.
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Answer 1

-2

Explanation:

Given equation is x²+px+1=0

comparing the equation with ax²+bx+c=0

then a=1,b=p,c=1

let the roots one root of the equation be A then

by question, another root is A².

we know

A(ALPHA)+A²(beta)=-b/a

A+A²=-p/1

A+A²=-p......1

again

A*A²=c/a

A³=1/1

A³=1

A=1

Now,

substituting A=1 in equation 1

1+1²=-p

2=-p

Therefore p=-2.