3 square roots in side each other equation :

Question

$$\sqrt{25-(x+1)\sqrt{4-(x-11)\sqrt{1+(x-6)(x-8)}}} = \frac{(9x-18-x^2)}{2} $$ Find (x)

I tried to take the square to the both sides but i couldn't continue,, can you help me ?? Thanks in advance

Answer 1

Just try calculating: Start with the term

$1+(x-6)(x-8)$

and see if you can take the square root of that and so on.

Answer 2

Firstly, note that $9x-18-x^2\ge 0 \quad \implies x \in [3,6]$

Then, start from the most bottom square root and try factoring it (and hoping it can be square root-ed). Make sure it is positive in $[3,6]$, so that the constraint is satisfied.

For example $\sqrt{1+(x-6)(x-8)} = \sqrt{(x-7)^2}=|x-7|=7-x$ (as $x-7 \ngeq 0 \quad\forall x\in[3,6])$

And then continue by tackling the next square root in the same manner.

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$\sqrt{25-(x+1)\sqrt{4-(x-11)\sqrt{1+(x-6)(x-8)}}} \\ =\sqrt{25-(x+1)\sqrt{4-(x-11)\sqrt{(x-7)^2}}}\\ =\sqrt{25-(x+1)\sqrt{4-(x-11)(7-x)}}\\ =\sqrt{25-(x+1)\sqrt{(x-9)^2}}\\ =\sqrt{25-(x+1)(9-x)}\\ =\sqrt{(x-4)^2}\\ =|x-4|$

(as we don't know whether $x\ge4$ or $x<4$)