# For any real number c, the quadratic equation $x^2+x-c^2 = 0$ has two distinct (real) solutions. Is this true or false and explain why.

I am a first year math major taking the introductory proofs course. This is my solution to the question. I would like you to check if my solution is correct or complete.

The statement is true.

In order for the quadratic equation to have two distinct real solutions. The discriminant has to be greater than 0. So, $$b^2-4ac \gt 0$$. I will change the c in the discriminant formula to d. In order to avoid confusion. $$b^2-4ad$$.

I know that b = 1, a = 1 and d = $$-c^2$$

So, = $$(1)-4(1)(-c^2)$$

= $$1+4c^2$$
$$\gt0$$

It is true because $$1+4c^2$$ will always be greater then 0. $$c^2\gt0$$ and $$1+4c^2$$ will never be
zero because of the 1 being added.

• Remember that $a=1$. Otherwise everything else looks fine. Sep 12, 2020 at 17:35
• It is not a good idea to assign two different meanings to the same variable, $c$. You said "$c=c^2$", but the "$c$" on the left-hand side is not the same as the "$c$" on the right-hand side. Sep 12, 2020 at 17:40
• Yes, bravo man. You got it right. The solution as you mentioned. But as Geoffrey mentioned, you should use another notation for c. Sep 12, 2020 at 17:43
• You want to show $b^2-4ac \gt 0$, not $\sqrt{b^2-4ac} \gt 0$. Remember, $\sqrt{x} \geq 0$ is always true by definition (for real $x \geq 0$ that is, for $x<0$ you would get into complex numbers and the inequality would make even less sense).
– Sil
Sep 12, 2020 at 17:49
• You wrote $c^2>0$, but in general only $c^2\ge0$. In the end, this changes nothing though as we add $1$ Sep 12, 2020 at 19:46

$$b^2$$-4ac= 1-4(1)(-$$c^2$$)=1+4$$c^2$$