Solve $\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $

Question

Solve $$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$

Preface; I think there should be an algebraic method of solving this equation for $x$ since graphing these two graphs

We get whole number solutions such as $x=0,2,3$

So I think there is someway of manipulating this equation into a disguised quadratic somehow!

So my attempt is this:

$$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$

Let $u=x^2-5x+8$ and $u-2=x^2-5x+6$ which means we can rewrite our equation as

$$\left(\sqrt{\sqrt{u}+\sqrt{u-2}} \right)^x + \left(\sqrt{\sqrt{u}-\sqrt{u-2}} \right)^x = 2^{\frac{x+4}{4}} $$

Squaring both sides we get

$$ \left( \sqrt{u} + \sqrt{u-2}\right)^x + 2\left(\sqrt{\sqrt{u}+\sqrt{u-2}} \right)^x\left(\sqrt{\sqrt{u}-\sqrt{u-2}} \right)^x+ \left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{\frac{x+4}{2}}$$

$$\left( \sqrt{u} + \sqrt{u-2}\right)^x +\left( \sqrt{u} - \sqrt{u-2}\right)^x+ 2(\sqrt{2})^x+ = 2^{\frac{x+4}{2}}$$

Now some little algebra

$2^{\frac{x+4}{2}}-2(\sqrt{2})^x=2^{\frac{x}{2}} \cdot 2^2 - 2 \cdot 2^{\frac{x}{2}}=2^{\frac{x}{2}} \cdot 2=2^{\frac{x+2}{2}}$

$$ \left( \sqrt{u} + \sqrt{u-2}\right)^x +\left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{\frac{x+2}{2}} $$

Square both sides again

$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x}+2\left( \sqrt{u} + \sqrt{u-2}\right)^x\left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{x+2} $$

$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x}+2(2^x) = 2^{x+2} $$

$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x} = 2^{x+1} $$

Now I've hit a roadblock.. :(

Answer 1

hint: Use AM-GM inequality: $a + b \ge 2\sqrt{ab}$, for this type of question, with $a,b$ are the first and second terms of the left side of the equation. The product $ab = $ constant. Specifically, with $a = \sqrt{\sqrt{x^2-5x+8} + \sqrt{x^2-5x+6}}, b = \sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}}$, we have equation occurs when $a = b \implies \sqrt{x^2-5x+6} = 0 \implies x = 2,3$.

Answer 2

HINT:

Replace $\sqrt u-\sqrt{u-2}=\dfrac2{\sqrt u+\sqrt{u-2}}$ to form a Quadratic equation in $$\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}$$ which is

$$\left(\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}-1\right)^2=0$$