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Question: Find the value of x, if \(x=\sqrt{6+\sqrt{6+\sqrt{6+.........\infty }}}\) where x is a natural numbe...

Find the value of x, if x=6+6+6+.........∞x=\sqrt{6+\sqrt{6+\sqrt{6+.........\infty }}} where x is a natural number.

Explanation

Solution

Hint: As x=6+(6+6+......∞)x=\sqrt{6+\left( \sqrt{6+\sqrt{6+......\infty }} \right)} so we can take the bracket part as x because this is also the infinite series so the above expression can be written as x=6+xx=\sqrt{6+x}. Now, squaring both the sides will give the quadratic in x and then solve for x.
Complete step-by-step answer:
It is given in the question that:
x=6+(6+6+......∞)x=\sqrt{6+\left( \sqrt{6+\sqrt{6+......\infty }} \right)}
Now, take the bracket part as x. We can take the bracket part as x because in the bracket also the series is infinite and removing something from the infinite won’t make a difference. So, the above expression will look as follows:
x=6+(x)x=\sqrt{6+(x)}
Squaring both the sides will give:
x2=6+x ⇒x2−x−6=0  \begin{aligned} & {{x}^{2}}=6+x \\\ & \Rightarrow {{x}^{2}}-x-6=0 \\\ & \\\ \end{aligned}
The above equation is a quadratic in x. We are going to solve the quadratic equation using factorization method:
x2−3x+2x−6=0 ⇒x(x−3)+2(x−3)=0 ⇒(x+2)(x−3)=0 \begin{aligned} & {{x}^{2}}-3x+2x-6=0 \\\ & \Rightarrow x\left( x-3 \right)+2\left( x-3 \right)=0 \\\ & \Rightarrow \left( x+2 \right)\left( x-3 \right)=0 \\\ \end{aligned}
As we can see from the above that 2 solutions are obtained for x.
x = 2 and x = -3
One solution is positive and the other is negative. Now, we know that a square root of a number cannot be negative. The x we have solved is equal to the square root of something so x cannot take the negative value. Hence, the solution x = -3 is rejected. And the above equation has the only solution of x which is 2.
Hence, the value of x is 2.
Note: After solving any polynomial equation, always check the solutions that you are getting are satisfying the parent equation or not.