Equation exercise asked by Ali Kalf in ‎اساتذة الرياضيات في ذي

Exercise:

Solve in $\mathbb{R}$ , ${{\left( \sqrt{2}-1 \right)}^{x}}+{{\left( \sqrt{2}+1 \right)}^{x}}-2\sqrt{2}=0$

Solution: we have $\sqrt{2}-1=\frac{1}{\sqrt{2}+1}$ since $\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$

So ${{\left( \frac{1}{\sqrt{2}+1} \right)}^{x}}+{{\left( \sqrt{2}+1 \right)}^{x}}-2\sqrt{2}=0\Leftrightarrow \frac{1}{{{\left( \sqrt{2}+1 \right)}^{x}}}+{{\left( \sqrt{2}+1 \right)}^{x}}=2\sqrt{2}$

$\Leftrightarrow 1+{{\left( \sqrt{2}+1 \right)}^{2x}}=2\sqrt{2}{{\left( \sqrt{2}+1 \right)}^{x}}$ Put $w={{\left( \sqrt{2}+1 \right)}^{x}}\Leftrightarrow 1+{{w}^{2}}=2\sqrt{2}w$

${{w}^{2}}-2\sqrt{2}w+1=0\Leftrightarrow {{w}^{2}}-2\sqrt{2}w+2-2+1=0$

$\Leftrightarrow {{\left( w-\sqrt{2} \right)}^{2}}-1=0\Leftrightarrow w-\sqrt{2}=\pm 1\Leftrightarrow w=\sqrt{2}\pm 1$

If $w=\sqrt{2}+1\Leftrightarrow {{\left( \sqrt{2}+1 \right)}^{x}}=\sqrt{2}+1\Leftrightarrow x=1$

If $w=\sqrt{2}-1\Leftrightarrow {{\left( \sqrt{2}+1 \right)}^{x}}=\sqrt{2}-1=\frac{1}{\left( \sqrt{2}+1 \right)}={{\left( \sqrt{2}+1 \right)}^{-1}}\Leftrightarrow x=-1$