Question

if alpha and beta are the zeroes of the quadratic polynomial p(x)=4x^2-5x-1,find the value of alpha^2×beta+alpha×beta^2

Answer 1

firstly solve it by middle term splitting method then in the expression of alpha2beta+alphabeta2 take alpha and beta as common

Answer 2

$\bf{\underline{\underline{Question:-}}}$

$\sf if\: \alpha,\beta \:are\:the\:zeroes\:of\:a\: Quadratic\: polynomial$

$\sf \:f(x) = 4x^2-5x-1, \:Find\:the\:value\:of\: \alpha^2\beta +\alpha \beta^2$

$\bf{\underline{\underline{Given:-}}}$

• $\sf Zeroes\:of\:the\: polynomial\: are\: \alpha,\beta$
• $\sf f(x) = 4x^2-5x-1$

$\bf{\underline{\underline{To\: Find:-}}}$

• $\sf Find\:the\:value\:of\: \alpha^2\beta +\alpha \beta^2$

$\bf{\underline{\underline{Solution:-}}}$

$\sf f(x) = 4x^2-5x-1$

Here,

• a = 4
• b = -5
• c = -1

$\sf → \alpha+\beta=\dfrac{-b}{a}$

$\sf → \alpha+\beta = -( \dfrac{-5}{4})$

$\sf→ \alpha+\beta = \dfrac{5}{4}$

Now,

$\sf →\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)$

$\sf → \dfrac{-1}{4}×\dfrac{5}{4}$

$\sf → \dfrac{-5}{16}$

$\bf{\underline{\underline{Hence:-}}}$

• $\sf The\: Required\:value\:of \alpha^2\beta+\alpha\beta^2\:is\: \dfrac{-5}{16}$