Question

The number of quadratic equations which remain unchanged by squaring their roots, is:

• A. 0
• B. 2
• C. 4
• D. infinitely many

Answer 1

The correct option is C

4

Let $\alpha ,\beta$ be the roots of a quadratic equation and ${\alpha }^2,{\beta }^2$ be the roots of another quadratic, since the quadratic remains the same, we have
$\alpha +\beta ={\alpha }^2+{\beta }^2\dots \left(i\right)and\alpha \beta ={\alpha }^2{\beta }^2\dots \left(ii\right)Now,\alpha \beta ={\alpha }^2{\beta }^2\Rightarrow \alpha \beta \ast (\alpha \beta -1)=0\Rightarrow \alpha =0or\beta =0or\alpha \beta =1$
If $\alpha =0$, then $\beta ={\beta }^2$ [putting $\alpha =0$ in(i)]
$\Rightarrow \beta (1-\beta )=0\Rightarrow \beta =0,\beta =1$
Thus, we get two sets of values of $\alpha$ and $\beta$ viz. $\alpha =0,\beta =0$ and $\alpha =0,\beta =1$. Now if $\alpha \beta =1$,
$\begin{array}{ccc}then& \alpha +\frac{1}{\alpha }={\alpha }^2+\frac{1}{{\alpha }^2}& [putting\beta =\frac{1}{\alpha }in(i\left)\right]\\ \Rightarrow & \alpha +\frac{1}{\alpha }={(\alpha +\frac{1}{\alpha })}^2-2& \\ \Rightarrow & {(\alpha +\frac{1}{\alpha })}^2-(\alpha +\frac{1}{\alpha })-2=0& \\ \Rightarrow & \alpha +\frac{1}{\alpha }=2or\alpha +\frac{1}{\alpha }=-1& \\ \Rightarrow & \alpha =1or\alpha =\omega ,{\omega }^2& \end{array}$
Putting $\alpha =1$, in $\alpha \beta =1$, we get $\beta =1$, and putting $\alpha =\omega$ in $\alpha \beta =1$, we get $\beta ={\omega }^2$
Putting $\alpha ={\omega }^2$ in $\alpha \beta =1$, we get $\beta =\omega$ , thus, we get four sets of values of $\alpha ,\beta$ viz., $\alpha =0,\beta =0;\alpha =0,\beta =1;\alpha =\omega ,\beta ={\omega }^2;\alpha =1,\beta =1$.
Thus, there are four quadratics which remain unchanged by squaring their roots.