Algebra Math Problem Find the exact value of x
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