# 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. – player3236 Sep 12 '20 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. – Geoffrey Trang Sep 12 '20 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. – Daddy Sep 12 '20 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 '20 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 $1$ – Hagen von Eitzen Sep 12 '20 at 19:46

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