Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-2x+4=0$, then what is the equation whose roots are $\frac{\alpha ^{3}}{\beta ^{2}}$ and $\frac{\beta ^{3}}{\alpha ^{2}}?$

Option 1: $x^{2}-4x+8=0$

Option 2: $x^{2}-32x+4=0$

Option 3: $x^{2}-2x+4=0$

Option 4: $x^{2}-16x+4=0$

Correct Answer: $x^{2}-2x+4=0$

Solution : Given: $\alpha$ and $\beta$ are roots of the equation $x^{2}-2x+4=0$
⇒ $\alpha+\beta=2$ and $\alpha\beta=4$
Now we need to find the equation whose roots are $\frac{\alpha ^{3}}{\beta ^{2}}$ and $\frac{\beta ^{3}}{\alpha ^{2}}$.
$\frac{\alpha^{3}}{\beta^{2}}+\frac{\beta^{3}}{\alpha^{2}}$
= $\frac{\alpha ^{5}+\beta^{5}}{\alpha^{2}\beta ^{2}}$
= $\frac{(\alpha ^{2}+\beta^{2}) \times(\alpha ^{3}+\beta^{3})-(\alpha ^{2}\beta ^{3}+\alpha ^{3}\beta ^{2})}{(\alpha\beta)^{2}}$
= $\frac{[((\alpha+\beta)^{2}-2\alpha\beta)\times((\alpha+\beta)^{3}-3\alpha\beta\times(\alpha+\beta))-\alpha^{2}\beta ^{2}\times(\alpha+\beta)]}{(\alpha\beta)^{2}}$
= $\frac{[((2)^{2}-2\times 4)\times((2)^{3}-3\times 4\times 2)-4^{2}\times(2)]}{(4)^{2}}$
= $\frac{((-4)\times(-16)-32)}{16}$
= $\frac{(64-32)}{16}$
= $\frac{32}{16}$
= $2$
Also, $\frac{\alpha^{3}}{\beta^{2}}×\frac{\beta^{3}}{\alpha^{2}}=\alpha\beta=4$
So, the required equation is $x^{2}-(\frac{\alpha^{3}}{\beta^{2}}+\frac{\beta^{3}}{\alpha^{2}})x+(\frac{\alpha^{3}}{\beta^{2}}×\frac{\beta^{3}}{\alpha^{2}})=0$
⇒ $x^2-2x+4=0$
Hence, the correct answer is $x^2-2x+4=0$.