Question

IF α,β,and gamma are the zeroes of the polynomials f(x)=px³+qx²+rx+s , then find the value of α²+β²+gamma²

Answer 1

$f(x) = p {x}^{3} + q {x}^{2} + rx + s$

If a,b,c are the roots of the above polynomial then a+b+c =>-q/p and ab+bc+ca=>r/p.

${(a + b + c)}^{2} = \frac{ {q}^{2} }{ {p}^{2} } = > {a}^{2} + {b}^{2} + {c}^{2} +$

$2ab + 2bc + 2ca = \frac{ {q}^{2} }{ {p}^{2} } = > {a}^{2} +{b}^{2} +$

${c}^{2} + \frac{2r}{p} = \frac{ {q}^{2} }{ {p}^{2} } = > {a}^{2} + {b}^{2} + {c}^{2} = \frac{ {q}^{2} }{ {p}^{2} } -$

$\frac{2r}{p} = > {a}^{2} + {b}^{2} + {c}^{2} = \frac{ {q}^{2} - 2pr}{ {p}^{2}}$

Hope it helps you.