I am stuck with this equation, I need to prove the roots are real when $a, b, c \in R$
The equation is $$(a+b-c)x^2+ 2(a+b)x + (a+b+c) = 0$$
If someone could tell me the right way to go about this, so I can attempt it.
Thank you
EDIT: I have made an error in the question. I have now corrected it.
Best Answer
We look at the discriminant of the the polynomial, which for a quadratic $ax^2 +bx +c$ is $b^2 -4ac$, plugging the values in for our polynomial gives $$\Delta = 4(a+b)^2-4(a+b-c)(a+b+c)\\ = 4[(a+b)^2 - (a+b)^2+c^2]\\ = 4c^2$$
Since the square of a real number is positive, we know that the roots must be real, by looking at the quadratic forumla and seeing that the solutions are $$\frac{-b\pm\sqrt\Delta}{2a}$$
And the square root of a positive real number is real. We used the discriminant because it makes computation so much easier, than if we were doing everything that we did in the first step underneath the radical, and it would be rather ugly. Inspection shows that if $\Delta > 0$, there are two distinct real roots, if $\Delta < 0$, there are two complex roots, which are conjugate, and if $\Delta = 0$ then you have a real double root.