Question

Let $f\left(x\right)$ be a quadratic expression such that $f(0+f(1)=0$. If $f(-2)=0$, then

• A. $f\left(\frac{-2}{5}\right)=0$
• B. $f\left(\frac{2}{5}\right)=0$
• C. $f\left(\frac{-3}{5}\right)=0$
• D. $f\left(\frac{3}{5}\right)=0$

Answer 1

Answer: (D)

$f\left(\frac{3}{5}\right)=0$

Let the quadratic expression be, $f\left(x\right)=a{x}^2+bx+c$

$\begin{array}{}f\left(\right(0)+f(1)=0⟹a+b+2c=0& \text{(1)}\end{array}$

$\begin{array}{}f(-2)=0⟹4a-2b+c=0& \text{(2)}\end{array}$

$2\times \left(1\right)+\left(2\right)$, we get $6a+5c=0⟹c=\frac{-6a}{5}$

$2\times \left(2\right)-\left(1\right)$, we get $7a-5b=0⟹b=\frac{7a}{5}$

$\therefore f\left(x\right)=a{x}^2+\frac{7a}{5}x-\frac{6a}{5}$

$⟹f\left(x\right)=a(x^2+\frac{7x}{5}-\frac{6}{5})$

$⟹f\left(x\right)=a(x-\frac{3}{5})(x+2)$

$\therefore f\left(\frac{3}{5}\right)=0$