Question

if the zeroes of the polynomial f(x) =2x3+3x2−9x−10 are x-y, x, x+y then the value of x is​

Answer 1

Appropriate Question :-

If $\bf f ( x ) = \sf 2x³ + 3x² - 9x - 10$ and $\quad \sf x - y \: , \: x \: , \: x + y \quad$ are zeroes of $\bf f ( x )$ . Find the value of x ?

Solution :-

Before Starting the question , let's recall vieta formula for a $\bf \orange{Cubic \: Polynomial}$ ;

Let ;

$\quad \qquad \dag \quad \bf \orange{p ( x )} = \sf ax³ + bx² + cx + d$

also

• If $\alpha$ , $\beta$ and $\gamma$ are the zeroes of $\bf \orange {p(x)}$ . Then we have ;

$\quad \qquad { \bigstar { \underline { \boxed { \pmb { \bf { \red { \underbrace { \alpha + \beta + \gamma = - \dfrac{b}{a} }}}}}}}}{\bigstar} \quad \qquad$

Using this , in case of $\bf f ( x )$ . We can say that ;

$\quad \leadsto \quad \sf ( x - y ) + x + ( x - y ) = - \dfrac{3}{2}$

${ : \implies \quad \sf x - y + x + x - y = - \dfrac{3}{2}}$

${ : \implies \quad \sf x - \cancel{y} + x + x - \cancel{y} = - \dfrac{3}{2}}$

${ : \implies \quad \sf 3x = - \dfrac{3}{2}}$

${ : \implies \quad \sf x = - \dfrac{3}{3 \times 2}}$

${ : \implies \quad \sf x = - \dfrac{\cancel{3}}{\cancel{3} \times 2}}$

${ : \implies \quad \bf \therefore \quad x = - \cfrac{1}{2}}$

Henceforth , The Required Answer is $\bf - \dfrac{1}{2}$