Help with solving a quadratic problem - rectangle inscribed within a circle

Question

I'm hoping someone can help me with the attached problem, from a high school textbook.

Please see my work below. I get stuck at part d. Can someone help explain how I should use the discriminant and the inequality at this point?

Thank you!

Answer 1

Quadratic from part (c) as per your working is correct.

$\displaystyle 2x^2 - bx + \frac{b^2}{4} - 4a^2$

For real roots, we must have $32a^2 - b^2 \geq 0 \implies b \leq 4\sqrt2 a$.

Please also note that the two roots of the equation are two sides of the rectangle and their sum should be $\geq 2a$ (triangular inequality $x + y \geq 2a$).

If there are two roots $r_1, r_2$ to the equation $ax^2+bx+c = 0$. Then using Vieta's formula, $\displaystyle r_1 + r_2 = - \frac{b}{a}$

So here, $\displaystyle r_1 + r_2 = \frac{b}{2} \geq 2a \implies b \geq 4a$.

That leads to $4a \leq b \leq 4\sqrt2 a$.