Find the value of $\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}} $

Question

Prepp Team · Accepted Answer

Understanding and Solving Infinite Nested Radicals

The problem asks us to find the value of the infinite nested radical expression: $\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}}$. This type of problem involves an expression that repeats infinitely within itself. We can solve such problems by using a simple algebraic technique.

Setting up the Equation for the Infinite Radical

Let the value of the given infinite nested radical expression be $x$.

So, we have:

$x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}}$

Observe the structure of the expression. The part under the first square root is $6$ plus the entire infinite expression itself. Since the expression repeats infinitely, the part $\sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}$ is exactly the same as the original expression we defined as $x$.

Therefore, we can substitute $x$ back into the expression under the first square root:

$x = \sqrt{6 + x}$

Solving the Equation to Find the Value

Now we have a simple equation involving $x$. To eliminate the square root, we can square both sides of the equation:

$x^2 = (\sqrt{6 + x})^2$

$x^2 = 6 + x$

This is a quadratic equation. To solve it, we rearrange the terms to set the equation equal to zero:

$x^2 - x - 6 = 0$

We can solve this quadratic equation by factoring. We need to find two numbers that multiply to -6 and add up to -1 (the coefficient of the $x$ term). These numbers are -3 and 2.

So, we can factor the quadratic equation as follows:

$x^2 - 3x + 2x - 6 = 0$

$x(x - 3) + 2(x - 3) = 0$

$(x - 3)(x + 2) = 0$

This gives us two possible values for $x$:

$x - 3 = 0 \implies x = 3$
$x + 2 = 0 \implies x = -2$

Choosing the Correct Value for the Nested Radical

We found two potential values for $x$: 3 and -2.

Let's consider the original expression: $\sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}}$. The square root symbol ($\sqrt{$}\) denotes the principal (non-negative) square root. Since the expression involves adding positive numbers (6) and taking square roots repeatedly, the result must be a non-negative number.

The value $x = 3$ is positive.
The value $x = -2$ is negative.

Therefore, the negative solution $x = -2$ is not valid in the context of the principal square root. The only valid solution is the non-negative one, which is $x = 3$.

Thus, the value of the infinite nested radical $\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}}$ is 3.

Summary of Solving the Infinite Nested Radical

To find the value of $\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}}$, we followed these steps:

Set the expression equal to a variable $x$.
Use the repeating nature to form the equation $x = \sqrt{6 + x}$.
Square both sides to get a quadratic equation: $x^2 = 6 + x$, which simplifies to $x^2 - x - 6 = 0$.
Solve the quadratic equation (by factoring in this case) to find the possible values for $x$: $x=3$ and $x=-2$.
Choose the non-negative solution because the principal square root must be non-negative.

The valid solution is $x=3$.

Step	Action	Equation/Result
1	Let the expression equal $x$	$x = \sqrt{6 + \sqrt{6 + \cdots}}$
2	Substitute $x$ into the repeating part	$x = \sqrt{6 + x}$
3	Square both sides	$x^2 = 6 + x$
4	Rearrange into quadratic form	$x^2 - x - 6 = 0$
5	Factor the quadratic	$(x - 3)(x + 2) = 0$
6	Find possible values for $x$	$x = 3$ or $x = -2$
7	Select the non-negative value	$x = 3$

Revision Table: Key Concepts

Concept	Explanation	Relevance to Problem
Infinite Nested Radical	A radical expression where the radicand contains the entire expression itself, repeating infinitely.	The core structure of the problem.
Substitution Method	Assigning a variable to the entire repeating expression to form an algebraic equation.	The primary technique used to convert the infinite expression into a solvable equation.
Quadratic Equation	An equation of the form $ax^2 + bx + c = 0$. Can be solved by factoring, completing the square, or the quadratic formula.	The resulting equation after squaring the radical equation.
Principal Square Root	The non-negative square root of a number. Represented by the symbol $\sqrt{$}\).	Crucial for determining the valid solution among the roots of the quadratic equation.

Additional Information: Infinite Expressions and Their Values

Infinite nested radicals are just one type of infinite expression in mathematics. Other examples include infinite series and infinite continued fractions.

Infinite Series: A sum of an infinite sequence of numbers, like $1 + \frac{1}{2} + \frac{1}{4} + \cdots$. Some infinite series converge to a finite value.
Infinite Continued Fractions: An expression of the form $a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cdots}}}$. These can also sometimes represent specific numbers, including irrational numbers like $\sqrt{2}$.

The method used here, setting the expression equal to a variable and using the self-referential nature to form an equation, is a common technique for evaluating expressions that have this kind of infinite, repeating structure, provided they converge to a finite value.

In the case of $\sqrt{c + \sqrt{c + \sqrt{c + \cdots}}}}$, if it converges, the value $x$ satisfies $x = \sqrt{c + x}$, leading to $x^2 - x - c = 0$. The positive solution of this quadratic gives the value of the nested radical.

Prepp Team · Answer

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Step	Action	Equation/Result
1	Let the expression equal \(x\)	\(x = \sqrt{6 + \sqrt{6 + \cdots}}\)
2	Substitute \(x\) into the repeating part	\(x = \sqrt{6 + x}\)
3	Square both sides	\(x^2 = 6 + x\)
4	Rearrange into quadratic form	\(x^2 - x - 6 = 0\)
5	Factor the quadratic	\((x - 3)(x + 2) = 0\)
6	Find possible values for \(x\)	\(x = 3\) or \(x = -2\)
7	Select the non-negative value	\(x = 3\)