This is the exciting part of quadratic equations. It is fun to see dancing radicals like this $\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3\ldots}}}}$. This is an unending chain of radical expression and can be surprisingly simplified using quadratic equations.

Worked Problem 1:

$\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+\ldots}}}}$ can be simplified in the form of $\displaystyle\frac{p+\sqrt{q}}{r}$. Find the value of $p+q+r$

Solution:

Let $x=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+\ldots}}}}$

By squaring both sides;

$x^2=(\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+\ldots}}}})^2$

$x^2=3+\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+\ldots}}}}$

Since $x=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+\ldots}}}}$, by substitution

$x^2=3+x$ or $x^2-x-3=0$

Since we are looking for x we can use quadratic formula to solve for it.

$x_1x_2=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{4a}$

$x_1x_2=\displaystyle\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-3)}}{4(1)}$

$x_1x_2=\displaystyle\frac{1\pm\sqrt{13}}{2}$

Only keep the positive x.

$x=\displaystyle\frac{1+\sqrt{13}}{2}$

Therefore, $p+q+r=1+13+2=16$

Worked Problem 2:

Simplify the radical $\sqrt{5\sqrt{5\sqrt{5\sqrt{5\ldots}}}}$.

Solution:

Let $x=\sqrt{5\sqrt{5\sqrt{5\sqrt{5\ldots}}}}$

By squaring both sides

$x^2=(\sqrt{5\sqrt{5\sqrt{5\sqrt{5\ldots}}}})^2$

$x^2=5\sqrt{5\sqrt{5\sqrt{5\sqrt{5\ldots}}}}$

Since $x=\sqrt{5\sqrt{5\sqrt{5\sqrt{5\ldots}}}}$ by substitution

$x^2=5x$ or $x^2-5x=0$

By factoring

$x(x-5)=0$

$x=0$ and $x=5$

Worked Problem 3:

Find x if . What is the value of x?

Solution:

Since  by substitution

$x^2=2$

$x=\sqrt{2}$