Question

if alpha and beta are the zeros of the quadratic polynomial f(x)= 6x^2+x-2, then find the value of alphabeta^2 + alpha^2beta

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

Let p(x) be the given polynomial function

Given Polynomial,

$\large{ \tt{p(x) = 6 {x}^{2} + x - 2}}$

Note

• Sum of Zeros : - x coefficient / x² coefficient

• Product of Zeros : constant term/ x² coefficient

Let $\alpha$ and $\beta$ be the zeros of given p(x)

Here,

Sum of Zeros :

$\large{ \alpha + \beta = - \frac{1}{6} }$

Product of Zeros :

$\large{ \alpha \beta = - \frac{2}{6} }$

Now,

$\alpha { \beta }^{2} + { \alpha }^{2} \beta \\ \\ \rightarrow \: \alpha \beta ( \alpha + \beta ) \\ \\ \rightarrow \: ( - \frac{1}{6} )( - \frac{ 2}{6} ) \\ \\ \rightarrow \: \frac{2}{36} \\ \\ \huge{ \rightarrow \ \underline{ \boxed{ \frac{1}{18} }}}$

Hence,the value of $\alpha { \beta }^{2} + { \alpha }^{2} \beta$ is 1/18

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Polynomial :- $f (x) = 6 {x}^{2} + x -2$

Number of zero :- 2 (that is $\alpha\: and \:\beta$)

Value of :- (alpha)(beta^2) + (alpha^2)(beta)

We know that,

Product of zeros = $\dfrac{c}{a}$

And sum of zeros = $\dfrac {-b}{a}$

So, purring in the values :-

Product of zeros = $\dfrac{- 2}{6}$...(1)

And sum of zeros = $\dfrac{-1}{6}$...(2)

Now, we need to find the value of :-

(alpha)(beta^2) + (alpha^2)(beta)

It can be written as :-

(alpha)(beta) [alpha + beta]

From (1) and (2),

$=( \dfrac{- 2}{6}$) ($\dfrac{-1}{6})$

$\bold {= \dfrac{1}{18}}$

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