
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

Re: Quadratic equation
$\displaystyle x^2+axb=0$
In order for there to be $\displaystyle 1$ repeated root, the discriminant of the quadratic has to be $\displaystyle 0$.
Can you remember how to calculate the discriminant?

Re: Quadratic equation
an alternate approach:
given that we have just one solution:
$\displaystyle x^2 + ax  b = (x  r)^2 = 0$
multiplying the square out, we get:
$\displaystyle x^2 + ax  b = x^2  2rx + r^2$
thus $\displaystyle a = 2r, b = r^2$.
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 $\displaystyle r^2$, but a is only in terms of r. so let's square a:
$\displaystyle a^2 = (2r)^2 = 4r^2$.
that's almost the "opposite of b" we just need to multiply b by 4:
$\displaystyle 4b = 4r^2$.
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:
$\displaystyle a^2 + 4b = 4r^2  4r^2 = 0$
(and we didn't even have to remember what the discriminant is, or use that horrid quadratic formula)