A relation involving solutions of a quadratic equation

Question

Consider the following quadratic equation: $$x^2 - (a + d) x + (a d - bc) = 0,$$ where $a, b, c, d \geq 0,$ and let $x_1 > 0, x_2 < 0$ be the solutions. I read in a book that since $$x^2 - (a + d) x + (a d - bc) = \Big[x - \frac{a + d}{2}\Big]^2 - \Big[ \frac{(a - d)^2}{4} + ad - bc \Big],$$ and $\frac{a + d}{2}$ is nonnegative, then $x_1 > |x_2|$. Can someone explain me why is it true?

Answer 1

Hint Since we know the roots of the given quadratic (and that it is monic), we can also write it as $$(x - x_1) (x - x_2) .$$

Additional hint On the other hand, expanding this expression and comparing like terms in $x$ in our two different ways of writing the polynomial gives $$a + d = x_1 + x_2 .$$