Question

If $r, s$ are non zero roots of $a_0+a_1 x+a_2 x^2=0\left(a_0, a_1, a_2 \in R\right.$ and $\left.a_2 \neq 0\right)$, then the equality $a_0+a_1 x+a_2 x^2=a_0\left(1-\frac{x}{r}\right)\left(1-\frac{x}{s}\right)$ holds

• A. for all values of $x, a_0 \neq 0$
• B. only when $x=0$
• C. only when $x = r$ or $x = s$
• D. only when $x=r$ or $x=s, a_0 \neq 0$

Answer 1

Answer: (A)

We have $r+s=-\frac{a_1}{a_2} ; \quad r s=\frac{a_0}{a_2}$
Simplifying the R.H.S. of the equation, we have
$a_0\left(1-\frac{x}{r}\right)\left(1-\frac{x}{s}\right) =a_0\left(1-\frac{x}{r}-\frac{x}{s}+\frac{1}{r s} x^2\right)=a_0\left(1-\left(\frac{r+s}{r s}\right) x+\frac{1}{r s} x^2\right) $
$ =a_0\left(1+\frac{a_1}{a_0} x+\frac{a_2}{a_0} x^2\right)=a_0+a_1 x+a_2 x^2$
Thus, the answer is (A) for all values of $x, a_0 \neq 0$