Question

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

• 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: (B)

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

${\alpha }^2-2\alpha +3=0$ and ${\beta }^2-2\beta +3=0$

Now, ${\alpha }^3-3{\alpha }^2+5\alpha -2$

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

Putting ${\alpha }^2-2\alpha =-3$ in above step,

$=\alpha (-3-\alpha +5)-2$

$=2\alpha -{\alpha }^2-2$

Again putting $-{\alpha }^2+2\alpha =3$ in above step,

$=3-2$

$=1$

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

Similarly, here we are substituting ${\beta }^2=2\beta -3$

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

$\beta (2\beta -3-\beta +1)+5$

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

$=2$ $[\because {\beta }^2-2\beta =-3]$

Hence, the required equation is given by $(x-1)(x-2)=0$

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

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