Question

If $\alpha ,\beta$ be the root of $x^2-3x+4=0$, then the equation whose roots are ${\alpha }^3-4{\alpha }^2+7\alpha -3$ and ${\beta }^3-2{\beta }^2+\beta +6$ is

• A. $x^2-3x+2=0$
• B. $x^2-2x+2=0$
• C. $x^2-x+1=0$
• D. $x^2-2x+1=0$

Answer 1

The correct option is A $x^2-3x+2=0$

Since, $\alpha ,\beta$ are the roots of $x^2-3x+4=0$
${\alpha }^2-3\alpha +4=0$
$\Rightarrow {\alpha }^3-3{\alpha }^2+4\alpha =0$ ..... $\left(i\right)$
Similarly, ${\beta }^3-3{\beta }^2+4\beta =0$ ..... $\left(ii\right)$
One root of the required equation is
${\alpha }^3-4{\alpha }^2+7\alpha -3$
$=3{\alpha }^2-4\alpha -4{\alpha }^2+7\alpha -3$ ...... From $\left(i\right)$
$=-{\alpha }^2+3\alpha -3$
$=-({\alpha }^2-3\alpha )-3$
$=-(-4)-3=1$
The other root is
${\beta }^3-2{\beta }^2+\beta +6$
$=3{\beta }^2-4\beta -2{\beta }^2+\beta +6$ ...... From $\left(ii\right)$
$={\beta }^2-3\beta +6$
$=-4+6=2$
$\therefore$ Required equation is

$(x-1)(x-2)=0$
i.e. $x^2-3x+2=0$
Hence, option 'A' is correct.