maths
posted by shyam .
Find the sum of the squares of the three solutions of the equation x3+3x2−7x+1=0.

Assume the equation has three roots a,b and c:
then
(xa)(xb)(xc)=0
x³(a+b+c)x²+(ab+bc+ca)xabc=0
which means
(a+b+c)=3 (negative of coeff.of x²)
(ab+bc+ca)=7 (coeff. of x)
Hence
a²+b²+c²
=(a+b+c)²2(ab+bc+ca)
=3²2(7)
=9(14)
=23