Thread: Quadratic equation

Hey!

Someone please answer the following question, I tried everything but I still couldn't do it right...

Given that the equation x^2+ax=b, where a and b are real numbers, has a unique solution, prove that a^2+4b=0

I would very much appreciate it if someone can show me how it is done.

Thanks in advance

In order for there to be repeated root, the discriminant of the quadratic has to be .

Can you remember how to calculate the discriminant?

an alternate approach:

given that we have just one solution:



multiplying the square out, we get:



thus .

now, we want to end up with a condition on a and b, but we don't know what they are. but if we work in terms of r, maybe we can come up with something.

well b is in terms of , but a is only in terms of r. so let's square a:

.

that's almost the "opposite of b" we just need to multiply b by 4:

.

now, r only exists (and the equations for a and b involving r), because we are assuming we have just one solution.

but IF WE DO, then it is clear that:



(and we didn't even have to remember what the discriminant is, or use that horrid quadratic formula)