# How would I solve $x^2-4x=y^2-4y$ without knowing the answer beforehand?

The equation is $x^2-4x=y^2-4y$ in the case where $x\ne y$. The answer is $x+y=4$.

I can start from $x+y=4$ and create the equation very easily, and I can substitute $x+4=y$ into the equation and show both sides are equal easily. I just don't get how I would find the answer if I didn't know it before hand, and all I had was the equation? Any advice?

• I just realized, assuming that x and y aren't equal, then x must equal +/- (y-4) and y must equal +/- (x-4), otherwise there's no way x(x-4)=y (y-4). Nov 27, 2017 at 2:11
• Be weary of the line of reasoning; it’s fallacious. Just because $ab = cd$ and $a \neq b$, doesn’t mean $a = c$. That trick only works when we have something like $ab = 0$, whence we can conclude $a$ or $b$ is zero. See the most recent answer to this question for the right application of this idea. It’s subtle. Nov 28, 2017 at 1:49

\begin{align} x^2 - 4x &= y^2 - 4y \\ x^2 - y^2 &= 4x - 4y \\ (x-y)(x+y) &= 4(x-y) \\ x+y &= 4\end{align} where dividing by $x-y$ is allowed since $x \neq y$.

• It seems so obvious now, thanks for showing the way! I wasn't seeing it at all. Nov 26, 2017 at 21:17

As an alternative, the solution that struck me first was completing the square in both $x$ and $y$. This is common when dealing with quadratics, especially once there are no $xy$ cross-terms. $$x^2 - 4x = y^2 - 4y$$ $$x^2 - 4x + 4 = y^2 - 4y + 4$$ $$(x-2)^2 = (y-2)^2$$ This means that either $x-2 = y-2$ or $x-2 = -(y-2)$, which means either $x = y$ or $x+y = 4$, as desired.

• I actually saw that I could complete the square when I was working on it. I missed that it would progress towards the solution though. Thanks for the alternative approach! Nov 27, 2017 at 1:38
• You're welcome. Look into quadratic forms and conic sections for more general problems of this type. Nov 27, 2017 at 1:44
• This is the way I first went, so +1 ;) Nov 27, 2017 at 4:37

Let $c$ be the common value of $x^2-4x$ and $y^2-4y$. Then $x$ and $y$ are both roots of the polynomial $t^2-4t-c$. Since we are assuming $x$ and $y$ are distinct, they are all of the roots, so $t^2-4t-c$ factors as $(t-x)(t-y)$. Since $(t-x)(t-y)$ expands to $t^2-(x+y)t+xy$, comparing the coefficients of $t$ gives $x+y=4$.

(Conversely, if $x+y=4$, then since $x$ and $y$ are both roots of $(t-x)(t-y)=t^2-(x+y)t+xy=t^2-4t+xy$, $x^2-4x$ and $y^2-4y$ are both equal to $-xy$.)

• Very interesting approach, but where did the -4t come from in your polynomial? Nov 27, 2017 at 2:04
• I'm not sure what you mean. Since $x^2-4x=c$, $x^2-4x-c=0$, and similarly for $y$. So $x$ and $y$ are roots of $t^2-4t-c$. Nov 27, 2017 at 2:32
• Okay so it comes from the original equation, I see now Nov 27, 2017 at 2:43
• And you're using the t for clarities sake I think? Nov 27, 2017 at 2:45

Write $x+y=d$. Then we have $y=d-x$ and so: $$(d-x)^2-4(d-x) = x^2-4x$$ so $$d^2-2dx-4d =-8x$$thus $$d(d-2x)-4(d-2x)=0$$ so $$(d-2x)(d-4)=0$$ If $d=2x$ we get $x=y$ which is impossible. So $d=4$ and we have $x+y=4$.

I take the opportunity of this question to consider a more general issue.

Here is a methodology that can be applied to any equation of the form :

$$f(x_1)=f(x_2) \tag{1}$$

where $$f$$ is a function from $$I$$ (subset of $$\mathbb{R}$$) to $$J:=f(I)$$.

(I intentionally take variables $$(x_1,x_2)$$ instead of $$(x,y)$$)

Of course, $$(x,x), \ x \in I$$ is a natural (trivial) set of solutions.

Once the study of variations of $$f$$ has been done, operate a segmentation of the range $$J$$ of $$f$$ into the following components : $$J=J_1 \cup J_2 \cup J_3 \cup \cdots$$

where $$J_k$$ is defined as the subset of $$J$$ of all $$y$$ that $$f(x)=y$$ has $$k$$ distinct solutions. As a consequence, each $$J_k$$ will generate $$k(k-1)$$ non-trivial solutions to equation (1) ; for example, when $$k=3$$, we will have solutions :

$$(x_1,x_2), (x_1,x_3), (x_2,x_3) \ \text{and, symmetrically} \ (x_2,x_1), (x_3,x_1), (x_3,x_2)$$

Remark : the "trivial" cases correspond to $$J_1$$.

Let us consider the following example :

$$y=f(x)=x(x-1)(x-2)(x-3) \ \ \text{on} \ \ I=[0, +\infty) \ \text{and} \ J=[-1,+\infty).$$

with the following curve, possessing in particular a local maximum in $$(-\tfrac12,a:=\tfrac{9}{16})$$

We will have

• $$J_1=(a,+\infty)$$ generating "trivial solutions" to (1),

• $$J_2=\{0,a\}$$ (set with two elements), generating sets of two non-trivial solutions,

• $$J_3=(0,a)$$ (a whole line segment !), generating sets of $$3 \times 2$$ solutions,

• $$J_4=(-1,0)$$ generating sets of $$6 \times 2$$ solutions.

Now, let us return to the case of function $$y=f(x)=x^2-4x$$ (see Fig. 2) :

we will clearly have :

• $$J_1={y=-4}$$ corresponding to point $$x=2$$ giving trivial solution $$(x_1,x_1)=(2,2)$$

• $$J_2=(-4,+\infty)$$, yielding 2 solutions $$(x_1,x_2)$$ and $$(x_2,x_1)$$ obtained by solving, for each $$y \in (-4,+\infty)$$ the quadratic equation :

$$x^2-4x=y$$

(take care, once again : this $$y$$ has nothing to do with the $$y$$ of the initial question $$f(x)=f(y)$$ !)

giving the two solutions :

$$x_{1,2}=2 \pm \sqrt{4+y^2}\tag{2}$$

A first remark is that

$$x_1+x_2=4 \ \ \ \text{which is the solution "as presented"}. \tag{3}$$

A second and final remark is that (3) describes exactly all solutions because $$x_1$$ (and the same for $$x_2$$) can take any value : for every value of $$x_1$$, one can find a value of $$y$$ such that (2) is fulfilled.