Problem 71 - A Graphical Approach to College Algebra

Question

Solve each equation involving "nested" radicals for all real solutions analytically. Support your solutions with a graph. $$\sqrt{\sqrt{x}}=x$$

Step-by-Step Solution

Answer

The real solutions are \( x = 0 \) and \( x = 1 \).

Step 1: Understand the Problem

The equation involves nested radicals where one radical is placed inside another. We need to find the value of \( x \) such that \( \sqrt{\sqrt{x}} = x \) is true.

Step 2: Simplify the Equation

To simplify the equation, we express both sides in terms of powers. Recall that \( \sqrt{x} = x^{1/2} \) and \( \sqrt{\sqrt{x}} = (\sqrt{x})^{1/2} = x^{1/4} \). Therefore, the equation \( \sqrt{\sqrt{x}} = x \) can be rewritten as \( x^{1/4} = x \).

Step 3: Solve the Simplified Equation

Considering the simplified form \( x^{1/4} = x \), raise both sides to the power of 4 to eliminate the fractional exponent: \((x^{1/4})^4 = x^4\). This simplifies to \( x = x^4 \).

Step 4: Solve the Polynomial Equation

Rearrange the equation \( x = x^4 \) as \( x^4 - x = 0 \). Factor the equation: \( x(x^3 - 1) = 0 \). This gives the solutions \( x = 0 \) or \( x^3 = 1 \).

Step 5: Find All Real Solutions

Solve \( x^3 = 1 \) by taking the cube root of both sides: \( x = 1 \). Thus the real solutions to the equation \( \sqrt{\sqrt{x}} = x \) are \( x = 0 \) and \( x = 1 \).

Step 6: Verify the Solutions

Substitute \( x = 0 \) into the original equation: \( \sqrt{\sqrt{0}} = 0 = 0 \). Substitute \( x = 1 \): \( \sqrt{\sqrt{1}} = 1 = 1 \). Both solutions satisfy the original equation.

Step 7: Graphical Support

Graph the functions \( y = \sqrt{\sqrt{x}} \) and \( y = x \) to visually verify the solutions. The graphs intersect at \( (0, 0) \) and \( (1, 1) \), supporting that \( x = 0 \) and \( x = 1 \) are the solutions.

Key Concepts

Equation Solving

Solving equations often means finding the values that satisfy a mathematical statement. With nested radicals, the challenge increases slightly due to the layers of square roots involved. Consider the equation \( \sqrt{\sqrt{x}} = x \). To tackle it, we first convert the radical expressions into exponential forms, which sometimes make the algebra more straightforward. Here, \( \sqrt{x} \) can be written as \( x^{1/2} \), and thus \( \sqrt{\sqrt{x}} \) translates into \( x^{1/4} \). The equation then becomes \( x^{1/4} = x \). This type of expression may initially look intimidating, but unraveling the layers through exponentiation simplifies the process. Raising both sides to the fourth power removes the fractional exponent and results in the simpler polynomial \( x = x^4 \). This process transforms the nested radical problem into a more familiar algebraic equation, easing the path to solutions.

Polynomial Equations

Once you translate a problem involving nested radicals into a polynomial equation, you shift the challenge to solving for the roots of this polynomial. For our equation \( x = x^4 \), rewriting it as \( x^4 - x = 0 \) highlights the polynomial structure. Factoring polynomials often helps to expose the solutions. In this case, you start by factoring out \( x \), resulting in \( x(x^3 - 1) = 0 \). This factorization gives two routes for finding solutions: either \( x = 0 \), or \( x^3 - 1 = 0 \), which further simplifies to \( x^3 = 1 \). Solving \( x^3 = 1 \) reveals that \( x = 1 \) is a solution. Therefore, we end up with \( x = 0 \) and \( x = 1 \) as the solutions for this specific nested radical equation. Recognizing and converting these situations into polynomial forms is crucial as it enables us to apply algebraic techniques like factoring to find solutions effectively.

Graphical Verification

Graphical verification serves as an excellent tool to confirm the solutions of an equation. In the context of \( \sqrt{\sqrt{x}} = x \), plotting the functions \( y = \sqrt{\sqrt{x}} \) and \( y = x \) on the Cartesian plane offers visual evidence of where these functions intersect, confirming the solutions we derived algebraically. By graphing, we notice that the two curves meet at the points \( (0, 0) \) and \( (1, 1) \). These intersection points visually verify the solutions \( x = 0 \) and \( x = 1 \), thus proving our analytical work correct. Graphs provide additional verification by highlighting where the curves align, reinforcing our analytical approach with visual evidence.