Question

If $a,b$ are the roots of a equation of a quadratic equation $x^2-3x+5=0$ then the equation whose roots are $(a^2-3a+7)$ and $(b^2-3b+7)$ is-

• A. $x^2+4x+1=0$
• B. $x^2-4x-1=0$
• C. $x^2+4x+4=0$
• D. $x^2-4x+4=0$

Answer 1

Answer: (D)

$x^2-4x+4=0$

a,b are roots of $x^2-3x+5=0$

$\therefore a^2-3a+5=0$

$b^2-3a+5=0$

Let $f\left(x\right)$ be equation with roots $\alpha =a^2-3a+7$ & $\beta =b^2-3b+7$

$\alpha =a^2-3a+7$$\therefore \alpha =2(\because a^2-3a+5=0)$

$\beta =b^2-3b+7$

$\therefore \beta =2(\because b^2-3a+5=0)$

$\therefore f\left(x\right)=(x-2)(x-2)$

$f\left(x\right)=x^2-4x+4=0$

$x^2-4x+4=0$ is required equation with roots

$a^2-3a+7$ & $b^2-3b+7$