Question

Let a, $\beta$ are the roots of the equation $x^2-2x+3=0$ then the equation whose roots are $P={3a}^2+5a-2$ and $Q={\beta }^3{-\beta }^2+\beta +5$ is _______________.

• A. $x^2+3x+2=0$
• B. $x^2-3x+2=0$
• C. $x^2-3x-2=0$
• D. $-x^2-3x+9=0$

Answer 1

Answer: (B)

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

Let $\alpha ,\beta$ be the roots of equation, $x^2-2x+3=0$

According to the formula of finding root of quadratic equation $a{x}^2+bx+c=0$

$x=\frac{(-b\pm \sqrt{b^2-4ac})}{2a}$

Given: $a=1,b=-2,c=3.$

$x=\frac{(2\pm \sqrt{{(-2)}^2-4\times 3})}{2\times 1}$

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

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

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

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

Value of $\alpha =1+i\sqrt{2}$ and $\beta =1-i\sqrt{2}$

On substituting we get :-

To find equation whose roots are $p={\alpha }^3-3{\alpha }^2+5\alpha -2q={\beta }^3-{\beta }^2+\beta +5$

$p={\alpha }^3-3{\alpha }^2+5\alpha -2p={(1+i\sqrt{2})}^3-3{(1+i\sqrt{2})}^2+5(1+i\sqrt{2})-2p=1+2\sqrt{2}{i}^3+3\sqrt{2}i(1+i\sqrt{2})-3(1-2+2\sqrt{2}i)+5+5\sqrt{2}i-2p=1-2\sqrt{2}i+3\sqrt{2}i-6-3+6-6\sqrt{2}i+5+5\sqrt{2}i-2p=12-11+8\sqrt{2}-8\sqrt{2}p=1$

$q={\beta }^3-{\beta }^2+\beta +5q={(1-i\sqrt{2})}^3-{(1-i\sqrt{2})}^2+(1-i\sqrt{2})+5q=1-2\sqrt{2}{i}^3-3\sqrt{2}i(1-i\sqrt{2})-(1-2-2\sqrt{2}i)+1-\sqrt{2}i+5q=1+2\sqrt{2}i-3\sqrt{2}i-6+1+2\sqrt{2}i+1-\sqrt{2}i+5q=8-6+4\sqrt{2}-4\sqrt{2}q=2$

Hence equation whose roots are $p={\alpha }^3-3{\alpha }^2+5\alpha -2$ and $q={\beta }^3-{\beta }^2+\beta +5$ is : $(x-1)(x-2)=0$

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

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