Prove that $b^2\gt 4ac \implies ax^2+bx+c=0$ has two distinct real roots.

Question

How to do this in the method of a conditional proof? Assuming only $b^2\gt 4ac$ how to arrive at the conclusion without manipulating $ax^2+bx+c=0$ to yield the quadratic equation?

Answer 1

You're going to have to manipulate the equation at some point.

By completing the square and rearranging, you obtain the implication $$ax^2+bx+c=0 \quad \Rightarrow \quad x \in \left\{ \frac{-b + \sqrt{b^2-4ac}}{2a}, \frac{-b - \sqrt{b^2-4ac}}{2a} \right\}$$

By substituting those values into the equation you obtain the $\Leftarrow$ direction.

Thus we have the equivalence $$ax^2+bx+c=0 \quad \Leftrightarrow \quad x \in \left\{ \frac{-b + \sqrt{b^2-4ac}}{2a}, \frac{-b - \sqrt{b^2-4ac}}{2a} \right\}$$ without needing the assumption that $b^2>4ac$.

The assumption that $b^2 > 4ac$ tells you that the two solutions are real and distinct.

Answer 2

Hint: Write the given equation in the form $$a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}=0$$ for $a\neq 0$

Answer 3

Write the equation $ax^2 +bx+c=0$ as

$$(x+\frac{b}{2a})^2 + (\frac{c}{a} - \frac{b^2}{4a^2}) =0$$

which is easily verified by multiplying out.

So the quadratic has two real solutions iff

$$(x+\frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$$

has $2$ real solutions.

Which is the case iff the fraction on the RHS is $>0$ so iff $b^2-4c >0$. (Based ultimately on the fact that $y^2=d$ has two real solutions iff $d>0$ and that $4a^2 >0$ and so does not affect the sign of the fraction).