How to solve an equation one of its sides is an absolute function I want to solve this equation $$\sqrt{x^{2}-6x+9}=9-2x$$ and I know that this equation is the same as $$\sqrt{(x-3)^{2}}=9-2x \implies |x-3| =9-2x \implies 9-2x =\pm (x-3)$$ but when I solve it $x=4$ or $x = 6$ and $6$ doesn't satisfy the equation, what did I do wrong?
 A: Because $\lvert x-3\rvert=x-3\iff x\geqslant3$ and $\lvert x-3\rvert=-(x-3)\iff x\leqslant 3$. So, solve the equation $x-3=9-2x$ only in $[3,\infty)$ and solve the equation $-(x-3)=9-2x$ only in $(-\infty,3]$.
A: $x=6$ is an extraneous solution. 
The RHS evaluated when $x=6$ gives $-3$, but the square root on the LHS can only give a positive number.
A: One more option:
Declare $$9-2x \ge 0 ~~~(1)$$ and square both sides
$$x^2-6x+9=81+4*x^2-36x \Rightarrow x^3-10x+24=0 \Rightarrow (x-4)(x-6) =0 \Rightarrow x=6 ~ or ~ x=4.$$ As per (1), $x \le 9/2$, Hence $x=4$ is the only solution of your equation.
A: Let me add a word or more on what it means to solve an equation.
You get an equation presented to you, with the instruction to find all the numbers that satisfy it. Your procedure goes like this: You ask, “If there is such a number, what properties must it have?” Then if the equation is not an identity, you exclude infinitely many numbers as inconsistent with the equation. This is what, in advanced mathematics, is called the uniqueness proof.
But your work is not done yet. You must check which of these (presumably finitely many) numbers do in fact satisfy your equation. This is called the existence proof by mathematicians.
Comments three:
(1) When I was a student in school, back in a previous geological era, we were marked down if we did not do the second part of the argument, which we called simply “checking”.
(2) I now go to a local high-school to help in math classes. Students are never required to check, and I find this unacceptable.
(3) In much of mathematics, the important part of the question is the uniqueness. Anybody can see that a positive integer is a product of primes. But the strength of the Fundamental Theorem of Arithmetic is not that fact, but the uniqueness of the expression as product of primes.
