Algebra Math Problem

Algebra problem: If x is equal to the square root of 2 plus, square root of 2 plus, square root of 2 up to infinity, then find the value of x

x=\sqrt{2+\sqrt{2+\sqrt{2+......\mathrm{to} \space \infty}}}

x=??

Solution to This Algebra Math Problem

We Know

x=\sqrt{2+\sqrt{2+\sqrt{2+......\mathrm{to} \space \infty}}}

so

x=\sqrt{2+x}

Square both sides then

x^2=2+x

Rearrange this equation then

x^2-x-2=0

This is a quadratic equation, so

x=\dfrac{1\pm\sqrt{1^2-4\times1\times(-2)}}{2\times1}

x=\dfrac{1\pm\sqrt{1+8}}{2}

x=\dfrac{1\pm3}{2}

That is

x=2  \space \space \mathrm{or} \space \space x=-1

We can clearly see that x is positive, then

x = 2