Question

If alpha, beta, gamma r the zeros of polynomial a(x)3 + b(x)2 + cx + d, find (alpha)2 + (beta)2 +(gamma)2

Answer 1

Answer:

α²+β²+γ² = b²/a²  -  2c/a

              = ( b² - 2ac ) / a²

Step-by-step explanation:

The elementary symmetric polynomials in the roots are related to the coefficients of ax³+bx²+cx+d as follows:

α+β+γ = -b/a

αβ+βγ+γα = c/a

αβγ = -d/a

So...

α²+β²+γ² = ( α + β + γ )² - 2 ( αβ + βγ + γα )

              = ( -b/a )² - 2c/a

              = b²/a²  -  2c/a

              = ( b² - 2ac ) / a²

Answer 2

Here, Sum of roots = -b/a

Sum of product of two roots = c/a

Thank You.