• Oct 16th 2011, 05:08 AM
IBstudent
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.

• Oct 16th 2011, 05:10 AM
Quacky
\$\displaystyle x^2+ax-b=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?
• Oct 16th 2011, 07:03 AM
Deveno
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)