If r and s are the roots of $$x^2 - 3x + 2 = 0$$​, then the quadratic equation in x whose roots are $$r^2 + s^2$$ and $$(rs)^2$$ is:

Correct option is C

Given:

If r and s are roots of given equation =$$x^2 - 3x + 2 = 0$$​​

Formula Used:

For finding the roots we use:

Quadratic formula :$$x =\frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$​​

If $$\alpha$$​ and $$\beta$$​ are roots of a equation:

Then equation is given by : $$x^2 - (\alpha + \beta)x + (\alpha\beta) = 0$$​​

Solution:

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

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

​$$=\frac{ 3 \pm \sqrt{9 - 8}}{2}$$​​

​$$x= \frac{3+1}{2} or x =\frac{3-1}{2}$$​​

r= 2 and s= 1

​$$\alpha = r^2 + s^2 = (2)^2 + (1)^2 = 4+1 = 5$$​​

​$$\beta = (rs)^2= (2\times 1)^2 = 4$$​​

Quadratic equation = $$x^2 - (5+4)x + (5)(4) =0$$​​

​$$x^2 - 9x +20 = 0$$​​