x 2 - 20 = $\rm\sqrt{20+\sqrt{20+\sqrt{20+... infinite \ terms}}}$ , then what is x equal to ?

Question

x 2 - 20 =  $\rm\sqrt{20+\sqrt{20+\sqrt{20+... infinite \ terms}}}$ , then what is x equal to ?

Prepp Team · Accepted Answer

Solving an Equation with an Infinite Nested Square Root

The problem asks us to find the value of x given the equation:

$x^2 - 20 = \sqrt{20+\sqrt{20+\sqrt{20+... infinite \ terms}}}$

Let's first focus on the infinite nested square root expression on the right side of the equation. This type of expression is quite common in mathematics.

Evaluating the Infinite Nested Radical

Let the infinite expression be represented by a variable, say y:

$y = \sqrt{20+\sqrt{20+\sqrt{20+...}}}$

Because the expression goes on infinitely, the part under the first square root (starting from the second term) is the same as the original expression y. So, we can write:

$y = \sqrt{20+y}$

Now, we can solve this equation for y. To get rid of the square root, we square both sides:

$y^2 = (\sqrt{20+y})^2$

$y^2 = 20+y$

Rearrange the terms to form a quadratic equation:

$y^2 - y - 20 = 0$

We can solve this quadratic equation by factoring. We need two numbers that multiply to -20 and add up to -1. These numbers are -5 and 4.

$(y - 5)(y + 4) = 0$

This gives us two possible values for y:

$y - 5 = 0 \implies y = 5$
$y + 4 = 0 \implies y = -4$

Since y was defined as the square root of a number (and we usually consider the principal, or non-negative, square root in such contexts), y must be non-negative. Therefore, we take the positive value:

$y = 5$

Finding the Value of x

Now that we know the value of the infinite expression is 5, we can substitute this back into the original equation:

$x^2 - 20 = y$

$x^2 - 20 = 5$

Now, we solve for x. Add 20 to both sides of the equation:

$x^2 = 5 + 20$

$x^2 = 25$

Take the square root of both sides to find the value of x:

$x = \pm \sqrt{25}$

$x = \pm 5$

So, the possible values for x are 5 and -5.

Comparing with Options

The options provided are:

4
5
$\sqrt{5}$
$2\sqrt{5}$

Comparing our possible values for x (5 and -5) with the given options, we see that 5 is one of the options.

Thus, the value of x that satisfies the equation among the given options is 5.

Revision Table: Key Concepts

Concept	Description
Infinite Nested Radical	An expression where a square root contains another square root, and this pattern repeats infinitely, like $\sqrt{a+\sqrt{a+\sqrt{a+...}}}$.
Solving Infinite Radicals	Set the expression equal to a variable (e.g., $y$), then use the property of infinity ($\sqrt{a+y}=y$) to form an equation and solve for the variable.
Quadratic Equation	An equation of the form $ay^2+by+c=0$. Can be solved by factoring, completing the square, or using the quadratic formula.
Principal Square Root	For a non-negative number, the principal square root is the non-negative root. $\sqrt{25} = 5$, not -5, although $(-5)^2 = 25$.

Additional Information: Infinite Series and Equations

Infinite nested radicals are examples of expressions that can sometimes be evaluated to a finite value. The key is recognizing the self-referential nature of the infinite structure. The process involves setting the expression equal to a variable and then substituting the variable back into the expression under the outermost radical sign.

For a general infinite nested radical of the form $\sqrt{a+\sqrt{a+\sqrt{a+...}}}$, if we let $y = \sqrt{a+\sqrt{a+\sqrt{a+...}}}$, then $y = \sqrt{a+y}$. Squaring both sides gives $y^2 = a+y$, or $y^2 - y - a = 0$. The positive solution to this quadratic equation gives the value of the infinite radical.

In our specific problem, $a=20$, leading to $y^2 - y - 20 = 0$, which we solved to get $y=5$ (the positive root).

Equations involving variables under square roots, like $x^2 = 25$, can have multiple solutions (positive and negative). However, the context of the problem or the options provided often guides which solution is relevant or sought.