Find $x$ given $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$

Question

If $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$, then what is the value of $x$?

Is there an easy way to solve such equation, instead of squaring on both sides and replacing $\sqrt{x-2}$ with a different variable?

Answer 1

$$\sqrt{x+14-8\sqrt{x-2}} +\sqrt{x+23-10\sqrt{x-2}} = 3$$ $$\sqrt{(\sqrt{x-2}-4})^2+\sqrt{(\sqrt{x-2}-5)^2}=3$$ Then $$|\sqrt{x-2}-4|+|\sqrt{x-2}-5|=3$$ Let $\sqrt{x-2}\ge5 (x\ge27)$. Then

$$\sqrt{x-2}-4+\sqrt{x-2}-5=3$$ $$\sqrt{x-2}=6$$

Let $\sqrt{x-2}\le4 (2\le x\le18)$. Then

$$4-\sqrt{x-2}-\sqrt{x-2}+5=3$$ $$\sqrt{x-2}=3$$

Answer 2

It's not so bad; after you substitute $y=\sqrt{x-2}$ you will have $$\sqrt{y^2 - 8y + 16 } + \sqrt{y^2 - 10y+ 25 } = 3$$ and you should be able to take it from here without any squaring of both sides.

Answer 3

Condition: $x\geq 2$.

You have $$LHS = \sqrt{(\sqrt{x-2}-4)^2} + \sqrt{(\sqrt{x-2}-5)^2}.$$

So, if $\sqrt{x-2} < 4$, or $x<18$, then $$LHS = 4-\sqrt{x-2} + 5-\sqrt{x-2} = 9-2\sqrt{x-2} = 3 = RHS.$$

If $4 \leq \sqrt{x-2} < 5$, or $18\leq x < 27$, then $$LHS = \sqrt{x-2} -4 + 5-\sqrt{x-2} = 1 \neq 3 = RHS.$$

If $\sqrt{x-2} \geq 5$, or $x \geq 27$, then $$LHS = \sqrt{x-2} - 4+ \sqrt{x-2} -5= 2\sqrt{x-2} - 9 = 3 = RHS.$$