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(A) Let the quadratic polynomial be $f(x) = a(x - \alpha)(x - \beta)$. Since one root is $-1$,let $\alpha = -1$. Then $f(x) = a(x + 1)(x - \beta)$.Giv

Vedclass · Accepted Answer

$\frac{11}{3}$

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(A) Let the quadratic polynomial be $f(x) = a(x - \alpha)(x - \beta)$. Since one root is $-1$,let $\alpha = -1$. Then $f(x) = a(x + 1)(x - \beta)$.Given $f(-2) + f(3) = 0$.$f(-2) = a(-2 + 1)(-2 - \beta) = a(-1)(-2 - \beta) = a(2 + \beta)$.$f(3) = a(3 + 1)(3 - \beta) = a(4)(3 - \beta) = a(12 - 4\beta)$.Summing these: $a(2 + \beta + 12 - 4\beta) = 0$.Since $a 
eq 0$,we have $14 - 3\beta = 0$,which gives $\beta = \frac{14}{3}$.The roots are $-1$ and $\frac{14}{3}$.The sum of the roots is $-1 + \frac{14}{3} = \frac{-3 + 14}{3} = \frac{11}{3}$.

Let $f(x)$ be a quadratic polynomial such that $f(-2) + f(3) = 0$. If one of the roots of $f(x) = 0$ is $-1$,then the sum of the roots of $f(x) = 0$ is equal to

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