Troubles with solving $\sqrt{2x+3}-\sqrt{x-10}=4$

Question

I have been trying to solve the problem $\sqrt{2x+3}-\sqrt{x-10}=4$ and I have had tons problems of with it and have been unable to solve it. Here is what I have tried-$$\sqrt{2x+3}-\sqrt{x-10}=4$$ is the same as $$\sqrt{2x+3}=4+\sqrt{x-10}$$ from here I would square both sides $$(\sqrt{2x+3})^2=(4+\sqrt{x-10})^2$$ which simplifies to $$2x+3=16+x-10+8\sqrt{x-10}$$ I would then isolate the radical $$x-3=8\sqrt{x-10}$$ then square both sides once again $$(x-3)^2=(8\sqrt{x-10})^2$$ which simplifies to $$x^2-6x+9=8(x-10)$$ simplified again $$x^2-6x+9=8x-80$$ simplified once again $$x^2-14x+89=0$$ this is where I know I have done something wrong because the solution would be $$14 \pm\sqrt{-163 \over2}$$ I am really confused and any help would be appreciated

Answer 1

Note that when you square something like $a\sqrt{b}$ you get $a^2b$.

Thus, you should get:

$\begin{align} x^2-6x+9 &= 64(x-10)\\ x^2-6x+9 &= 64x-640\\ x^2-70x+649 &= 0\\ (x-11)(x-59) &= 0\\ \therefore \boxed{x=11,59}. \end{align}$

Answer 2

Let $~t=\sqrt{x-10}.~$ Then $~x=t^2+10.~$ Replacing, we have $~\sqrt{2t^2+23}-t=4,~$ which can be

rewritten as $~\sqrt{2t^2+23}=t+4.~$ Squaring both sides, we have $~2t^2+23=t^2+8t+16.~$ Then,

subtracting, we are left with $~t^2-8t+7=0,~$ whose two roots are $~t=1~$ and $~t=7,~$ where

the former returns $~x=11,~$ while the latter yields $~x=59.~$ The reason why I wanted to share

this method with you is because, in my opinion, it is less confusing $~($and therefore less prone to

basic calculation errors$)~$ than the one you already tried.

Answer 3

Alternatively, take the reciprocal to get:

$$\frac{1}{\sqrt{2x + 3} - \sqrt{x - 10}} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{(2x+3)-(x-10)} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{x+13}$$

which we know is equal to $\frac 1 4$.

Thus we have the following:

$$\sqrt{2x+3} - \sqrt{x+10} = 4 \tag{1}$$ $$\sqrt{2x + 3} + \sqrt{x - 10} = \frac{1}{4}(x+13) \tag{2}$$

Add the two together to get:

$$2 \sqrt{2x + 3} = 4 + \frac{1}{4}(x + 13)$$ $$8\sqrt{2x+3} = 16 + (x + 13)$$ $$64(2x + 3) = x^2 + 58x + 841$$ $$x^2-70x+649=0$$ $$(x - 11)(x - 59) = 0 \implies x = \boxed{11, 59}.$$