Q92 / 155 · WBJEE 2016 · Mathematics
If $\alpha$ and $\beta$ are roots of $a x^{2}+b x+c=0,$ then
the equation whose roots are $\alpha^{2}$ and $\beta^{2},$ is
- Correct answerA$a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$
- B$a^{2} x^{2}+\left(b^{2}-a c\right) x+c^{2}=0$
- C$a^{2} x^{2}+\left(b^{2}+a c\right) x+c^{2}=0$
- D$a^{2} x^{2}+\left(b^{2}+2 a c\right) x+c^{2}=0$
Solution
Given. $\alpha$ and $\beta$ are roots of $a x^{2}+b x+c=0$
$\therefore \quad \alpha+\beta=\frac{-b}{a}$
and $\quad a \beta=\frac{c}{a}$
Now, if the roots are $\alpha^{2}$ and $\beta^{2}$, then
$$
\begin{aligned}
a^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\
&=\left(\frac{-b}{a}\right)^{2}-\frac{2c}{a}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a} \\
\text { And } \quad \alpha^{2} \beta^{2} &=(\alpha \beta)^{2}=\left(\frac{c}{a}\right)^{2}=\frac{c^{2}}{a^{2}}
\end{aligned}
$$
The equation whose roots are $\alpha^{2}$ and $\beta^{2}$. is
$$
x^{2}-\left(\frac{b^{2}}{a^{2}}-\frac{2 c}{a}\right) x+\frac{c^{2}}{a^{2}}=0
$$
$\therefore \quad a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$
Hence, the correct answer is option: 'A'.