Question

If we have the expression \[x=\sqrt{6+\sqrt{6+\sqrt{+6...\infty }}}\], then choose which of the options hold.
(a) x is an irrational number
(b) \[2(c) \[x=3\]
(d) None of these

Answer 1

Hint: To find the sum of given infinite series, substitute the value of infinite sum as the variable x itself, i.e., rewrite \[x=\sqrt{6+\sqrt{6+\sqrt{+6...\infty }}}\] as \[x=\sqrt{6+x}\]. Square the equation on both sides and factorize the given equation to find roots of the equation and then check each of the given options.

Complete step-by-step answer:
We have to find the value of \[x=\sqrt{6+\sqrt{6+\sqrt{+6...\infty }}}\].

As this is an infinite sum, we can write the terms of the equation as the variable x itself. Thus, we have \[x=\sqrt{6+x}\].

We have to simplify the given equation. To do so, we will square the equation on both sides.
Thus, we have \[{{x}^{2}}=6+x\].

Rearranging the terms of the above equation, we have \[{{x}^{2}}-x-6=0\].

We can rewrite the above equation as \[{{x}^{2}}+2x-3x-6=0\].

Taking out the common terms, we have \[x\left( x+2 \right)-3\left( x+2 \right)=0\].

So, we have \[\left( x+2 \right)\left( x-3 \right)=0\].

Thus, the roots of the above equation are \[x=-2,3\].

Now, we will check each of the given options.

We can clearly see that x is not an irrational number. Hence, option (a) is incorrect.

Also, x does not lie between 2 and 3. Hence, option (b) is incorrect as well.

We have \[x=3\] as one of the solutions of the above equation. Thus, this is a correct option.

Hence, the solution of the equation \[x=\sqrt{6+\sqrt{6+\sqrt{+6...\infty }}}\] is \[x=3\], which is option (c).

Note: We observe that the given equation when simplified is a polynomial equation. Polynomial is an expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication or division, and non-negative integer exponents of variables. Degree of a polynomial is the value of the highest power of degrees of its individual term. We observe that the polynomial given to us is of degree 2. There are multiple ways to solve a polynomial equation, like completing the square and factoring the polynomial by splitting the intermediate terms. We have solved this question using the factorization method by splitting the intermediate terms.