If f(x) is a quadratic expression such that f | vedclass

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(D) Let the quadratic expression be $f(x) = k(x - r_1)(x - r_2)$.Given that $-1$ is a root,let $r_1 = -1$. Let the other root be $a$.So,$f(x) = k(x +

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$\frac{8}{5}$

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(D) Let the quadratic expression be $f(x) = k(x - r_1)(x - r_2)$.Given that $-1$ is a root,let $r_1 = -1$. Let the other root be $a$.So,$f(x) = k(x + 1)(x - a) = k(x^2 + (1 - a)x - a)$.We are given $f(1) + f(2) = 0$.$f(1) = k(1 + 1)(1 - a) = 2k(1 - a) = 2k - 2ka$.$f(2) = k(2 + 1)(2 - a) = 3k(2 - a) = 6k - 3ka$.Summing these: $f(1) + f(2) = (2k - 2ka) + (6k - 3ka) = 8k - 5ka$.Setting the sum to zero: $8k - 5ka = 0$.Since $f(x)$ is a quadratic expression,$k 
eq 0$,so we can divide by $k$.$8 - 5a = 0 \Rightarrow 5a = 8 \Rightarrow a = \frac{8}{5}$.Thus,the other root is $\frac{8}{5}$.

If $f(x)$ is a quadratic expression such that $f(1) + f(2) = 0$,and $-1$ is a root of $f(x) = 0$,then the other root of $f(x) = 0$ is

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