1. **State the problem:** Prove that the quadratic equation $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a > 0$, has two real distinct solutions if and only if $b^2 > 4ac$. 2. **Recall the quadratic formula:** The solutions to $ax^2 + bx + c = 0$ are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Key concept:** The term under the square root, $\Delta = b^2 - 4ac$, is called the discriminant. 4. **Analyze the discriminant:** - If $\Delta > 0$, then $\sqrt{\Delta}$ is a positive real number, so there are two distinct real solutions. - If $\Delta = 0$, then $\sqrt{\Delta} = 0$, so there is exactly one real solution (a repeated root). - If $\Delta < 0$, then $\sqrt{\Delta}$ is not real, so there are no real solutions. 5. **Prove the "if" part:** Assume $b^2 > 4ac$ (i.e., $\Delta > 0$). Then the solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Since $\sqrt{b^2 - 4ac} > 0$, the two solutions differ by $$\frac{2\sqrt{b^2 - 4ac}}{2a} = \frac{\sqrt{b^2 - 4ac}}{a} \neq 0$$ Thus, two distinct real solutions exist. 6. **Prove the "only if" part:** Assume the equation has two distinct real solutions. Then the discriminant must be positive, $b^2 - 4ac > 0$, because if it were zero or negative, the solutions would be repeated or non-real. 7. **Conclusion:** The quadratic equation $ax^2 + bx + c = 0$ has two distinct real solutions if and only if $$b^2 > 4ac$$