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.
In the information
discriminant is
$b^2$-4ac= 1-4(1)(-$c^2$)=1+4$c^2$
Which is always >0 as c€R