Question

If alpha and beta are the zeros of the quadratic polynomial PX equal to 4 x square - 5 x minus 1 find the value of Alpha square plus Alpha into beta square

Answer 1

$\large \pmb{\mathfrak{\red{Appropriate~question...}}}$

If $\alpha$ and $\beta$ are the zeroes of the polynomial $P(x)=4x^2-5x-1$, then find the value of $\alpha^2\beta+\alpha\beta^2$.

$\large \pmb{\mathfrak{\red{Given...}}}$

Given an equation $P(x)=4x^2-5x-1$ whose zeroes are $\alpha$ and $\beta$.

$\large \pmb{\mathfrak{\red{To~find...}}}$

We have to find the value of $\alpha^2\beta+\alpha{\beta}^2$.

$\large \pmb{\mathfrak{\red{Solution...}}}$

Firstly, we know that $\alpha^2\beta+\alpha\beta^2$ can be also written as $\alpha \beta(\alpha+\beta)$.

Now, we'll find the sum and product of the zeroes :

$\red{\pmb{\mathfrak {Product~ of ~zeroes=\dfrac{c}{a}}}}$

$:\implies\alpha+\beta= \dfrac{-1}{4}$

$\red{\pmb{\mathfrak{ Sum~of ~zeroes=\dfrac{-b}{a}}}}$

$:\implies \alpha\beta=\dfrac{-(-5)}{4}$

$:\implies \alpha\beta=\dfrac{5}{4}$

Now, we'll substitute the sum and product in $\alpha\beta(\alpha+\beta)$ :

$:\implies \dfrac{-1}{4}(\dfrac{5}{4})$

$:\implies \dfrac{-5}{16}$

$\red{\pmb{\mathfrak{ Henceforth,~\alpha^2\beta+\alpha\beta^2=\dfrac{-5}{16}}}}$