Question

What is $$\sqrt {12 + \sqrt {12 + \sqrt {12 + \sqrt {12 + \sqrt {12} ....} } } } $$?

Answer 1

Hint: Here in this question, we have to equate the infinite nested root to a variable. For this, first the term inside the radical can be replaced with the variable then by taking the square and solving the variable by using a method of factorization, we get the required solution.

Complete step by step answer:
In algebra, a nested radical is a radical expression that contains (nests) another radical expression
Consider the given expression:
$$\sqrt {12 + \sqrt {12 + \sqrt {12 + \sqrt {12 + \sqrt {12} ....} } } } $$
Here, we have an infinite number of nested square roots. We can equate the whole thing to the variable x.
$$ \Rightarrow \,\,\,\,x = \sqrt {12 + \sqrt {12 + \sqrt {12 + \sqrt {12 + \sqrt {12} ....} } } } $$
The above expression is never ending, the term inside the 1st square root can be written as $$12 + x$$.
$$ \Rightarrow \,\,\,x = \sqrt {12 + x} $$
Now take the square root on both sides.
$$ \Rightarrow \,\,\,{x^2} = 12 + x$$
Take all the RHS terms to the LHS then We get a quadratic equation as follows,
$$ \Rightarrow \,\,\,{x^2} - x - 12 = 0$$
To get the value of x by solving the quadratic equation. We can solve the equation by factorization.
Now, Break the middle term as the summation of two numbers such that its product is equal to -12. Calculated above such two numbers are -4 and 3.
$$ \Rightarrow \,\,\,{x^2} - 4x + 3x - 12 = 0$$
Making pairs of terms in the above expression
$$ \Rightarrow \,\,\,\left( {{x^2} - 4x} \right) + \left( {3x - 12} \right) = 0$$
Take out greatest common divisor GCD from the both pairs, then
$$ \Rightarrow \,\,\,x\left( {x - 4} \right) + 3\left( {x - 4} \right) = 0$$
Take $$\left( {x - 4} \right)$$ common
$$ \Rightarrow \,\,\,\left( {x - 4} \right)\left( {x + 3} \right) = 0$$
Equate each factor with zero.
$$ \Rightarrow \,\,\,\left( {x - 4} \right) = 0$$ and $$\left( {x + 3} \right) = 0$$
$$ \Rightarrow \,\,\,x - 4 = 0$$ and $$x + 3 = 0$$
$$\therefore \,\,\,x = 4$$ and $$x = - 3$$
Therefore, $$x$$ can have 2 values $$x = 4, - 3$$.
But, according to the question, we have $$x$$ as the square root of some numbers. As square roots cannot be negative, $$x$$ also cannot be negative. So, we take the value $$x = 4$$ So, the required solution is 4.

Note:
 Remember the number inside the square root radical will always be positive. If the equation is a quadratic equation. This problem can be solved by using the quadratic formula as well. The factors for the equation depend on the degree of the equation.