Question

If α and β are the zeroes of p(x) =3x²-2x-6 , then find α³β²+α²β³​

Answer 1

GIVEN FUNCTION -

p(x) = 3 {x}^{2} - 2x - 6

IT HAVE TWO ROOTS -

\alpha \: \: and \: \: \beta

WE KNOW THAT DIFFERENCE OF ROOTS IS -

\alpha - \beta = \frac{ \sqrt{d} }{a}

\alpha - \beta = \frac{ \sqrt{ {b}^{2} - 4ac } }{a}

\alpha - \beta = \frac{ \sqrt{ {( - 2) }^{2} - 4(3)( - 6) } }{3}

\alpha - \beta = \frac{ \sqrt{78} }{3}

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