Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

(A) Yes,such a quadratic equation exists. Consider the quadratic equation $2x^2 + x - 4 = 0$. Here,the coefficients $2, 1,$ and $-4$ are rational numbers.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-4)}}{2(2)}$
$x = \frac{-1 \pm \sqrt{1 + 32}}{4}$
$x = \frac{-1 \pm \sqrt{33}}{4}$
The roots are $\frac{-1 + \sqrt{33}}{4}$ and $\frac{-1 - \sqrt{33}}{4}$. Since $\sqrt{33}$ is an irrational number,both roots are irrational.