Question

If $\alpha ,\beta$ are the roots of the equation $x^2-2x+3=0$, obtain the equation whose roots are ${\alpha }^3-3{\alpha }^2+5\alpha -2,{\beta }^3-{\beta }^2+\beta +5$.

• A. $x^2-3x+2=0$
• B. $x^2+3x-2=0$
• C. $-x^2-3x+2=0$
• D. $-x^2+3x-2=0$

Answer 1

Answer: (A), (D)

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

If $\alpha ,\beta$ are the roots of $x^2-2x+3=0$
then ${\alpha }^2-2\alpha +3=0$ ...(1)
and ${\beta }^2-2\beta +3=0$ ....(2)
$\therefore {\alpha }^2=2\alpha -3,{\alpha }^3=2{\alpha }^2-3\alpha$
$\therefore P=(2{\alpha }^2-3\alpha )-3{\alpha }^2+5\alpha -2$
$=-{\alpha }^2+2\alpha -2=3-2=1,$ by (1)
Similarly $Q=2\therefore S=3,P=2$
Hence reqd. eq. is $x^2-3x+2=0.$ or $-x^2+3x-2=0$