This is what my lecturer taught me.

If you have $\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}$

You can easily convert to $A+B=C+D$ or $AB=CD$

And then he gave me an example.

$\sqrt{8x+1}+\sqrt{3x-5}=\sqrt{7x+4}+\sqrt{2x-2}$ Find x.

Which is totally work.

After that, I've tried to make my own problem.

But there are some of the problems doesn't work with this technique.

For example, $\sqrt{5x+25}+\sqrt{x}=\sqrt{4x}+\sqrt{3x-5}$.

If I use this technique I'll get $x=30$.

But if I solve the problem normally by power 2 both side I'll get $x = \frac { 40 } { 11 } + \frac { 10 \sqrt { 115 } } { 11 }$.


Does anyone know how this technique works?

And what're the limitations?

  • 2
    $\begingroup$ This would work if one among $A$ and $B$ is $0$ $\textbf{and}$ one among $C$ and $D$ is $0$. OR in general $\sqrt{AB}=\sqrt{CD}$. $\endgroup$ – Yadati Kiran Jan 30 at 6:52
  • $\begingroup$ Yes, it will if (A or B is zero) and (C or D is Zero). But he told me you can use with any equation with 4 square root. And It usually works I've done that. But there are some equations doesn't. $\endgroup$ – b.ben Jan 30 at 6:59
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    $\begingroup$ I think it's not very sensible to think that you can just convert the equation with square roots to the equation without square roots, if it most of the time doesn't work. So I don't recommend trying to learn this. Instead, you can think when the equations are equivalent. The main point is that $$ (\sqrt{A} + \sqrt{B})^2 = |A| + 2\sqrt{AB} + |B| \neq A +B $$ $\endgroup$ – Matti P. Jan 30 at 7:00
  • $\begingroup$ @MattiP. Totally agreed I shouldn't learn this. I just figured out the example he gave if we recheck the answer we will see that A=C and B=D. $\endgroup$ – b.ben Jan 30 at 7:25

Try $A=25$, $B=1$ and $C=D=9$.

We see that $$A+B\neq C+D$$ and $$AB\neq CD,$$ but $$\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}.$$

We can try to understand, when it happens.

Firstly, $A$, $B$, $C$ and $D$ are non-negatives.

If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$

Now, let $A>B$, $C>D$, but $A+B=C+D$.

Thus, after squaring of the both sides we obtain $$A+B+2\sqrt{AB}=C+D+2\sqrt{CD},$$ which gives $$AB=CD.$$ Thus, $$(A+B)^2-4AB=(C+D)^2-4CD$$ or $$(A-B)^2=(C-D)^2$$ or $$A-B=C-D,$$ which after summing with $$A+B=C+D$$ gives $$A=C$$ and $$B=D.$$


Squaring both sides: $\sqrt A+\sqrt B=\sqrt C + \sqrt D\implies A+B+2\sqrt{AB}=C+D+2\sqrt{CD}$

Therefore the technique works if $A$ or $B=0$ and $C$ or $D=0$.

  • $\begingroup$ I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D. $\endgroup$ – b.ben Jan 30 at 7:21
  • $\begingroup$ If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $\sqrt{x}=\sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2\sqrt x=2\sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$ $\endgroup$ – klirk Jan 30 at 18:30

If $A+B=C+D$ and $AB=CD$, then $\sqrt{AB}=\sqrt{CD}$, so that $(\sqrt{A}+\sqrt{B})^2 = A+2\sqrt{AB}+B=C+2\sqrt{CD}+D=(\sqrt{C}+\sqrt{D})^2$.

So if you have an equation of the form $\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}$, you can form the equations $A+B=C+D$ and $AB=CD$, and try to find a simultaneous solution to both equations, and that will give a solution to the original equation.

It's not enough to find a solution to just one of $A+B=C+D$ and $AB=CD$, as others have pointed out.


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