Question

If $x=\sqrt{4}.\sqrt[4]{44}.\sqrt[8]{84}.\sqrt[16]{164}........\infty$, then

• A. $x^2-8x+16=0$
• B. $x^2-3x+2=0$
• C. $x^2-5x+4=0$
• D. $x^2+5x+4=0$

Answer 1

The correct option is A $x^2-8x+16=0$

$\begin{array}{c}x=4^{1/2}{4}^{1/4}{4}^{1/8}\cdots \infty \\ =4^{\frac{1}{2}+\frac{1}{4}+\cdots +\infty }\\ =4^{\frac{1/2}{1-\frac{1}{2}}}\\ =4\\ {\left(x-4\right)}^2=0\\ x^2+16-8x=0\end{array}$