Proving roots to be real

Question

Given :

$$(a-b)x^2 + (c-a)x + (a-b) = 0 \text{ where } a,b,c \in Q $$

How would I prove that the roots of the equation will be real

Answer 1

You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $\pm i$.

Answer 2

You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac \gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.

Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.