$a(x-a)^2+b(x-b)^2=0$ has one solution; $a, b$ are not $0$. Prove $|a|=|b|$ $a(x-a)^2+b(x-b)^2=0$ has one solution; $a, b$ are not $0$. Prove $|a|=|b|$
simplifying the equation I got:
$$(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$$
solving for the Discriminant $D=0$, I got: 
$$a(a^2b-2ab^2+b^3)=0$$
and since $a$ cannot equal $0$, I got:
$$a=b \implies |a|=|b|$$
In the same way I also got: $$b=a$$
But I don't need the absolute values of $a, b$ because I got $a=b$ (which means $|a|=|b|$ is not needed) and checking in the equation $a$ cannot equal $-b$ because then the equation will have infinitely many solutions instead of just one as specified in the task.
Now, because I need to prove that $|a|=|b|$ and not that $a=b$, I think that I am missing something. So, am I missing something or is my solution correct?
 A: Your solution is correct, since it eventually leads to $(a-b)^2 = 0$ which implies that $a=b$. It has a small error though, since you still need to consider the possibility of $A=2(a+b) = 0 \Leftrightarrow  a = -b$ which then implies $|a|=|b|$. For the rest, of course, it is :
$$a=b \Rightarrow |a| = |b|$$
Thus, in any case, it must be $|a| = |b|$.
Note the strict, one way implication $(\Rightarrow)$. This only holds since $a=b$. If it was $|a| = |b|$ then you cannot say it is $a=b$.
It asks you to prove that if the equation has one solution, then the following holds. Strictly mathematically this is an one way implication $(\Rightarrow)$ which truly leads to the desired result by your solution. Of course, if it was an iff case which is a two-way implication $(\Leftrightarrow)$ then it wouldn't hold.
A: Let $$f(x)=a(x-a)^2+b(x-b)^2$$
if $a=-b$ then
$$f(x)=-4xa^2$$
if the root is zero, then $a=-b$.
When you computed the discriminant $D$, you had to assume $a+b\ne 0$.
A: Oooh boy!  Incorporating hamam_Abdallah's answer into your efforts:
Note: you want $(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$ to have one solution.
So you naturally did the quadratic equation to get the solutions are 
$x= \frac {2(a^2 +b^2) \pm \sqrt{D}}{2(a+b)}$ and for that to have one unique solution you need $D= 0$.
That's fine and good up to the point that you assume $2(a+b) \ne 0$.
You forgot to take into account the possibility that $2(a+b) = 0$ or $a = -b$.
If $a = - b$ you get:
$(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0\implies -2(a^2+b^2)x+(a^3+b^3)=0$ which is a linear equation and has exactly one solution (assuming $-2(a^2 +b^2) \ne 0$ which it doesn't if $a=-b \ne 0$).
So for $(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$ to have one solution EITHER:
1)  $(a+b) = 0$ and the equation is a linear equation.
OR
2) $(a+b) \ne 0$ and $D = (-2(a^2 + b^2))^2 - 4(a+b)(a^3 + b^3) = 0$ and the equation is a quadratic with a double root.
If 1) then we get $a = -b$.
If 2) then we get $a = b$.
.....
(I didn't actually follow your calculations but mine got the same result:
$4(a^2 + b^2)^2 - 4(a+b)(a^3 + b^3) = 0$
$a^4 + 2a^2b^2 + b^4 = a^4 + ab^3 + a^3b + b^3$
$2a^2b^2 = ab^3 + a^3b$
$2ab = b^2 + a^2$
$(b -a)^2 = 0$
$b =a$.
)
===
FWIW.  
If 1) $a = -b$ then solution is $x = -\frac {a^3 + b^3}{-2(a^2 + b^2)}=\frac {a^3 - a^3}{-2(a^2 + a^2)} = 0$.
And if 2)$a = b$ then the solution is $x = \frac {2(a^2 + b^2)\pm \sqrt{D}}{2(a+b)} = \frac {2a^2}{2a} = a$.
