Question

What is the value of $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+........\infty }}}}}$?

Answer 1

Hint: Assume the given expression as x. Now, write the expression of the sum of infinite terms as $x=\sqrt{7+\sqrt{7-x}}$. Square both the sides and take the term 7 to the L.H.S while leaving the term $\sqrt{7-x}$ in the R.H.S. Again square both the sides and form a bi – quadratic equation in x. Find the two roots by the hit and trial method and then two more roots by the discriminant formula of a quadratic equation given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Here, a is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term. Reject the negative values of x.

Complete step by step solution:
Here we have been provided with the expression $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+........\infty }}}}}$ and we are asked to determine its value. Let us assume the value of this expression as x, so we have,
$\Rightarrow x=\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+........\infty }}}}}$
As we can see that there are infinite terms in the given expression so we cannot add them directly. Now, we can see that at a certain step the expression is repeating itself so we can write it as:
\[\Rightarrow x=\sqrt{7+\sqrt{7-x}}\]
On squaring both the sides we get,
\[\Rightarrow {{x}^{2}}=7+\sqrt{7-x}\]
Taking 7 to the L.H.S and again squaring both the sides we get,
\[\begin{align}
  & \Rightarrow {{\left( {{x}^{2}}-7 \right)}^{2}}=\left( 7-x \right) \\
 & \Rightarrow {{x}^{4}}+49-14{{x}^{2}}=7-x \\
 & \Rightarrow {{x}^{4}}-14{{x}^{2}}+x+42=0 \\
\end{align}\]
As we can see that the above expression is a bi – quadratic equation because the highest exponent of x is 4. Therefore, we have to apply the hit and trial method to get the first 2 roots of the equation. Substituting x = -2 in the above equation we get,
\[\Rightarrow 16-14\left( 4 \right)+\left( -2 \right)+42=58-58=0\]
Clearly x = -2 is a root of the equation that means x + 2 = 0 is a factor of this equation, so we can write the equation as:
\[\Rightarrow \left( x+2 \right)\left( {{x}^{3}}-2{{x}^{2}}-10x+21 \right)=0\]
Now, substituting x = 3 we get,
\[\Rightarrow \left( 3+2 \right)\left( 27-18-30+21 \right)=5\left( 48-48 \right)=0\]
Clearly x = 3 is another root of the equation so x – 3 = 0 is also a factor of the equation, so we can write the equation as:
\[\Rightarrow \left( x+2 \right)\left( x-3 \right)\left( {{x}^{2}}+x-7 \right)=0\]
On equating the quadratic equation with 0 and using the discriminant formula given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where a is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term, we get,
\[\begin{align}
  & \Rightarrow \left( {{x}^{2}}+x-7 \right)=0 \\
 & \Rightarrow x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( 1 \right)\left( -7 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-1\pm \sqrt{29}}{2} \\
\end{align}\]
Now, we can see that in the expression $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+........\infty }}}}}$ the final radical term is the positive square root so x cannot be negative. Therefore, we have to reject the negative values of x. Also, if we will substitute \[x=\dfrac{-1+\sqrt{29}}{2}\] it will not satisfy the equation \[x=\sqrt{7+\sqrt{7-x}}\] so three values of x are rejected.
Hence, the value of the expression is 3.

Note: Note that you have to satisfy the equation \[x=\sqrt{7+\sqrt{7-x}}\] and not \[{{x}^{2}}=7+\sqrt{7-x}\] because the latter one is not the original equation but the former one is the original. If you will try to satisfy the latter equation then you will see that $x=\dfrac{-1-\sqrt{29}}{2}$ will also be the answer and if you will consider the equation \[{{\left( {{x}^{2}}-7 \right)}^{2}}=\left( 7-x \right)\] then all the 4 values of x will be the answer. However, we will have only one answer so you must focus on the original equation.