If $x^2-3x+2$ is a factor of $x^4-a{x}^2+b,$ then the equation whose roots are $a$ and $b$ is

The remainder is zero of equation $x^2-3x+2$

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

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

$x\left(x-2\right)-1\left(x-2\right)=0$

$\left(x-2\right)\left(x-1\right)=0$

If $x-1=0$

$x=1$

Putting value of $x$ in equation $x^4-a{x}^2+b=0$ remainder be zero

${\left(1\right)}^4-a{\left(1\right)}^2+b=0$

$-a+b=-1...............\left(i\right)$

If,  $x-2=0$

$x=2$

Putting value of $x=2$

${\left(2\right)}^4-a{\left(2\right)}^2+b=0$

$-4a+b=-16.......................\left(ii\right)$

After solving there equations $\left(i\right) \& \left(ii\right)$, the value of $a=5, b=4$

Here, $\alpha =5 \& \beta =4$

Now, the new quadratic equation

$x^2-\left(\alpha +\beta \right)+\alpha .\beta =0$

$x^2-\left(5+4\right)x+5\times 4=0$

Hence, the equation whose  roots are $a \& b$ 

$x^2-9x+20=0$