When do quadratics have roots?

(r1)^2-(r2^2)=|r1||r1-r2| implies 2(r1)(r2)+(r2)^2=0 or r2=0 (since |r1|>|r2|) This implies a^2-4(b-c)=a^2 or b=c. But this leads to a contradiction, as it is told that b and c are distinct.

We proceed via proof by contradiction:

Assume on the contrary, that there is at least one equation $x^2+ax+b-c=0$ with roots $r_1, r_2 \in \mathbb{R}$ , and that all other conditions of this problem are satisfied. Then there exists a pair of real numbers $r_1$ and $r_2$ such that $(x-r_1 )(x-r_2 )=x^2-(r_1+r_2 )x+r_1 r_2=x^2+ax+b-c=0$ . By completing the square, we see that this implies that $\left(x+\dfrac{a}{2}\right)^2+b-c-\left(\dfrac{a}{2}\right)^2 = 0 \implies x=-\dfrac{a}{2} \pm \sqrt{\left(\dfrac{a}{2}\right)^2-b+c}$ .

Without loss of generality, take $r_1=-\dfrac{a}{2}-\sqrt{\left(\dfrac{a}{2}\right)^2-b+c}$ and $r_2=-\dfrac{a}{2}+\sqrt{\left(\dfrac{a}{2}\right)^2-b+c}$ . We see that $r_1^2-r_2^2 = \left[-\dfrac{a}{2}-\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right]^2 - \left[-\dfrac{a}{2}+\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right]^2 \\ = \left[ \left(\dfrac{a}{2}\right)^2 + a\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} + \left(\dfrac{a}{2}\right)^2-b+c \space \right] - \left[ \left(\dfrac{a}{2}\right)^2-a\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} + \left(\dfrac{a}{2}\right)^2 - b + c \space \right] \\ \therefore r_1^2-r_2^2 = 2a\sqrt{\left(\dfrac{a}{2}\right)^2-b+c}.$ Now we check the value of $|r_1 |⋅|r_1-r_2 |$ :

$\left| -\dfrac{a}{2}-\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right| \cdot \left| \left(-\dfrac{a}{2}-a\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right)-\left(-\dfrac{a}{2}+\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right) \right| \\ =\left| \left(-\dfrac{a}{2}-\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right) \left(-2\sqrt{\left(\dfrac{a}{2}\right)^2-b+c}\space \right) \right| \\ =\left| \left(a\sqrt{\left(\dfrac{a}{2}\right)^2-b+c}+2\left[\left( \dfrac{a}{2} \right)^2-b+c\right] \space \right) \right| \\ =\left(2\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right) \left| \dfrac{a}{2}+\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right|.$ Taking $r_1^2-r_2^2=|r_1 |⋅|r_1-r_2 |$ leaves us with the following: $2a\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} = \left(2\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right) \left| \dfrac{a}{2}+\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \space \right| \\ \therefore a = \dfrac{a}{2}+\sqrt{\left(\dfrac{a}{2}\right)^2-b+c} \implies a = 2\sqrt{\left(\dfrac{a}{2}\right)^2-b+c}.$

Of course, the equation $a=2\sqrt{\left(\dfrac{a}{2}\right)^2-b+c}$ can only make sense if $b = c$ , which contradicts the assumption that $b \ne c$ . Hence, the statement is always false.