Question

Evaluate:

$\sqrt{72+\sqrt{72+\sqrt{72+\sqrt{72+....\infty }}}}$

Answer 1

Step 1:

Form a quadratic equation:

Let,

$x=\sqrt{72+\sqrt{72+\sqrt{72+\sqrt{72+....\infty }}}}$

This implies that,

$x=\sqrt{72+x}$

Square both sides of the above equation.

$$\begin{gathered} x=\sqrt{72+x} \\ \begin{array}{rcl}{x}^2& =& {\left(\sqrt{72+x}\right)}^2\\ & \Rightarrow & x^2=72+x\\ & \Rightarrow & x^2-x-72=0\end{array} \end{gathered}$$

Step 2:

Solve the quadratic equation:

Use factorization method to solve the equation.

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & x^2-9x+8x-72=0\\ & \Rightarrow & x\left(x-9\right)+8\left(x-9\right)=0\\ & \Rightarrow & \left(x+8\right)\left(x-9\right)=0\end{array} \\ \begin{array}{rcl}& \Rightarrow & x+8=0\mathrm{or}x-9=0\\ & \Rightarrow & x=-8\mathrm{or}x=9\end{array} \end{gathered}$$

But the value under the square root cannot be negative, so $x=-8$ is rejected.

$\therefore x=9$

Hence, $\sqrt{72+\sqrt{72+\sqrt{72+\sqrt{72+....\infty }}}}$ is $9$.