Question

If $\alpha ,\beta$ are roots of $e{q}^n{x}^2-2x+3=0$ then find the an $e{q}^n$ whose roots are ${\alpha }^3-3{\alpha }^2+5\alpha -2,{\beta }^3-{\beta }^2+\beta +5$.

Answer 1

Given $\alpha$ & $\beta$ are the roots of equation $x^2-2x+3=0$

To find: Equation whose roots are ${\alpha }^3-3{\alpha }^2+5\alpha -2$,

${\beta }^3-{\beta }^2+\beta +5$

Sol: $x^2-2x+3=0$

$x=\frac{2\pm \sqrt{4-12}}{2}$

$x=\frac{2\pm \sqrt{8i}}{2}$

$x=1\pm \sqrt{2}i$

$\alpha =1+\sqrt{2}i$, $\beta =1-\sqrt{2}i$

${\alpha }^3-3{\alpha }^2+5\alpha -2=(1+\sqrt{2}i{)}^3-3(1+\sqrt{2}i{)}^2+5(1+\sqrt{2}i)-2$

$=1+2\sqrt{2}{i}^3+3\sqrt{2}i(1+\sqrt{2}i)-3(1+2i^2+2\sqrt{2}i)+5+5\sqrt{2}i-2$

$=1-2\sqrt{2}i+3\sqrt{2}i-6-3+6-6\sqrt{2}i+3+5\sqrt{2}i$

$=1+\sqrt{2}i-\sqrt{2}i$

$=1$

${\beta }^3-{\beta }^2+\beta +5=(1-\sqrt{2}i{)}^3-(1-\sqrt{2}i{)}^2+(1-\sqrt{2}i)+5$

$=1-(\sqrt{2}i{)}^3-3\sqrt{2}i(1-\sqrt{2}i)-(1+2i^2-2\sqrt{2}i)+(1-\sqrt{2}i)+5$

$=2$

$\therefore$ Equation whose roots are ${\alpha }^3-3{\alpha }^2+5\alpha -2$ and ${\beta }^3-{\beta }^2+\beta +5$ is $(x-1)(x-2)=0$

$\Rightarrow x^2-3x+2=0$

$\therefore$ $x^2-3x+2=0$.