Problem 75
Question
Solve each equation involving "nested" radicals for all real solutions analytically. Support your solutions with a graph. $$\sqrt[3]{\sqrt{32 x}}=\sqrt[3]{x+6}$$
Step-by-Step Solution
Verified Answer
The solutions are \( x = 2 \) and \( x = 18 \).
1Step 1: Simplify and Remove Cube Roots
Given the equation \( \sqrt[3]{\sqrt{32x}} = \sqrt[3]{x+6} \), the first step is to remove the cube roots by raising both sides to the power of 3. This will yield: \( \sqrt{32x} = x + 6 \).
2Step 2: Remove the Square Root
To further simplify \( \sqrt{32x} = x + 6 \), square both sides to eliminate the square root: \( 32x = (x + 6)^2 \).
3Step 3: Expand and Rearrange
Expand the right-hand side: \( 32x = x^2 + 12x + 36 \). Rearrange the equation into a standard quadratic form: \( x^2 - 20x + 36 = 0 \).
4Step 4: Solve the Quadratic Equation
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -20 \), and \( c = 36 \). Calculate the discriminant: \( (-20)^2 - 4 \times 1 \times 36 = 400 - 144 = 256 \). Therefore, \( x = \frac{20 \pm \sqrt{256}}{2} = \frac{20 \pm 16}{2} \). This gives solutions: \( x = 18 \) and \( x = 2 \).
5Step 5: Verify Solutions
Substitute \( x = 18 \) and \( x = 2 \) back into the original equation to verify. For \( x = 18 \): \( \sqrt[3]{\sqrt{32 \cdot 18}} = \sqrt[3]{18 + 6} \), simplifies to \( \sqrt[3]{24} = \sqrt[3]{24} \). For \( x = 2 \): \( \sqrt[3]{\sqrt{32 \cdot 2}} = \sqrt[3]{2 + 6} \), simplifies to \( \sqrt[3]{8} = \sqrt[3]{8} \). Both solutions satisfy the original equation.
6Step 6: Graphical Support
Graph the functions \( y = \sqrt[3]{\sqrt{32x}} \) and \( y = \sqrt[3]{x+6} \) on the same coordinate plane. Observe the intersection points at \( x = 2 \) and \( x = 18 \), which confirm the analytical solutions.
Key Concepts
Cube RootsSquare RootsQuadratic Equations
Cube Roots
Cube roots are operations that help us find a number which, when multiplied by itself thrice, gives the original number. They are typically represented using the radical symbol with a small 3, such as in \( \sqrt[3]{x} \). Understanding cube roots is essential to solve equations involving nested radicals, such as the one given in our problem.
To eliminate cube roots in an equation, we often raise both sides to the power of three. This operation simplifies the equation, making it easier to solve further down the line. For example, in the exercise, by cubing both sides of \( \sqrt[3]{\sqrt{32x}} = \sqrt[3]{x+6} \), we remove the cube roots, simplifying it to \( \sqrt{32x} = x + 6 \). This step is crucial as it transforms the original complex nested radical expression into a more straightforward form that can be worked with more conventionally.
When dealing with cube roots, it's vital to remember that they can have both negative and positive real numbers as solutions. Also, in equations involving cube roots, always check the domain to ensure that the solutions fit within the context of the problem.
To eliminate cube roots in an equation, we often raise both sides to the power of three. This operation simplifies the equation, making it easier to solve further down the line. For example, in the exercise, by cubing both sides of \( \sqrt[3]{\sqrt{32x}} = \sqrt[3]{x+6} \), we remove the cube roots, simplifying it to \( \sqrt{32x} = x + 6 \). This step is crucial as it transforms the original complex nested radical expression into a more straightforward form that can be worked with more conventionally.
When dealing with cube roots, it's vital to remember that they can have both negative and positive real numbers as solutions. Also, in equations involving cube roots, always check the domain to ensure that the solutions fit within the context of the problem.
Square Roots
Square roots are operations where we find a number that, when multiplied by itself, results in the original number. Square roots are denoted by the radical sign \( \sqrt{} \). In our problem, after removing the cube roots, we end up with \( \sqrt{32x} = x + 6 \).
To solve or simplify an equation involving square roots, we typically square both sides of the equation to eliminate the square root. Notice that squaring \( \sqrt{32x} = x + 6 \) results in \( 32x = (x + 6)^2 \).
When dealing with square roots, it's important to ensure that the expression under the radical (also known as the radicand) is non-negative, since the square root of a negative number isn't a real number. This constraint often provides insight into the limits on potential solutions. Always verify each potential solution by substituting it back into the original equation. This step ensures that no extraneous roots are introduced during the solving process.
To solve or simplify an equation involving square roots, we typically square both sides of the equation to eliminate the square root. Notice that squaring \( \sqrt{32x} = x + 6 \) results in \( 32x = (x + 6)^2 \).
When dealing with square roots, it's important to ensure that the expression under the radical (also known as the radicand) is non-negative, since the square root of a negative number isn't a real number. This constraint often provides insight into the limits on potential solutions. Always verify each potential solution by substituting it back into the original equation. This step ensures that no extraneous roots are introduced during the solving process.
Quadratic Equations
Quadratic equations are polynomial equations of degree 2, generally written in the form \( ax^2 + bx + c = 0 \). These are central when dealing with the kind of transformation we did in our exercise.
In the provided solution, we simplify the initial nested radical expression to \( 32x = (x+6)^2 \), and further to \( x^2 - 20x + 36 = 0 \), which is a standard quadratic equation.
To solve quadratic equations, one of the most common methods is using the quadratic formula:
Key points to remember about quadratic equations include the discriminant \( (b^2 - 4ac) \), which determines the nature of the roots. A positive discriminant indicates two distinct real roots, zero implies one real root (with multiplicity two), and a negative value would mean no real roots. Always check solutions for their relevancy and correctness in the original problem context.
In the provided solution, we simplify the initial nested radical expression to \( 32x = (x+6)^2 \), and further to \( x^2 - 20x + 36 = 0 \), which is a standard quadratic equation.
To solve quadratic equations, one of the most common methods is using the quadratic formula:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Key points to remember about quadratic equations include the discriminant \( (b^2 - 4ac) \), which determines the nature of the roots. A positive discriminant indicates two distinct real roots, zero implies one real root (with multiplicity two), and a negative value would mean no real roots. Always check solutions for their relevancy and correctness in the original problem context.
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