Problem when convert $\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}$ to A+B=C+D.

Question

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.

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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 $x = \frac { 40 } { 11 } + \frac { 10 \sqrt { 115 } } { 11 }$.

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Questions:

Does anyone know how this technique works?

And what're the limitations?

Answer 1

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 $$B=D.$$

Answer 2

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 $D=0$.

Answer 3

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.