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Chapter 8: Problem 61

Solve each equation. Check the solutions. $$x^{2 / 3}+x^{1 / 3}-2=0$$

Short Answer

Expert verified
The solutions are \(x = -8\) and \(x = 1\).

Step by step solution

01

Substitute variables

Let us start by making a substitution to simplify the equation. Let us set \(y = x^{1/3}\). This gives us the new equation: \[ y^2 + y - 2 = 0 \]
02

Factorize the quadratic

Now, factorize the quadratic equation \(y^2 + y - 2 = 0\). This can be rewritten as: \[ (y + 2)(y - 1) = 0 \] The solutions to this equation are \(y + 2 = 0\) and \(y - 1 = 0\).
03

Solve for y

Solve for \(y\) in each equation. We get: \[ y + 2 = 0 \Rightarrow y = -2 \] and \[ y - 1 = 0 \Rightarrow y = 1 \]
04

Substitute back in terms of x

Remember that \(y = x^{1/3}\). Substitute back to find \(x\). We have: \[ x^{1/3} = -2 \] and \[ x^{1/3} = 1 \] Now, solve for \(x\) in each case: For \(x^{1/3} = -2\): \[ x = (-2)^3 = -8 \]For \(x^{1/3} = 1\): \[ x = 1^3 = 1 \]
05

Verify solutions

Substitute \(x = -8\) and \(x = 1\) back into the original equation to verify they are solutions. For \(x = -8\): \[ (-8)^{2/3} + (-8)^{1/3} - 2 = 4 - 2 - 2 = 0 \]For \(x = 1\): \[ 1^{2/3} + 1^{1/3} - 2 = 1 + 1 - 2 = 0 \]Both solutions satisfy the original equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

substitution method
The substitution method is a valuable technique to simplify complex equations. In this exercise, we encounter a fractional exponent, which can be challenging to handle directly.
To simplify, we introduce a new variable. Let鈥檚 set \( y = x^{1/3} \). This transforms the original equation \( x^{2/3} + x^{1/3} - 2 = 0 \) into a standard quadratic equation \( y^2 + y - 2 = 0 \).
By substituting, we turn a hard problem into a much easier one. This step provides a clearer path to solve the equation, making calculations more manageable.
quadratic factorization
Next, we use quadratic factorization to solve the simplified equation. Quadratic equations have a standard form \( ax^2 + bx + c = 0 \). Here, it鈥檚 \( y^2 + y - 2 = 0 \).
Factorization means expressing the quadratic as a product of two binomials. Let鈥檚 rewrite our equation:
\( (y + 2)(y - 1) = 0 \).
To find the values of \( y \), we set each binomial equal to zero:
\( y + 2 = 0 \) implies \( y = -2 \)
\( y - 1 = 0 \) implies \( y = 1 \)
Factorization helps us break down complex expressions systematically, simplifying problem-solving processes.
verification of solutions
After obtaining solutions, always verify them to ensure they satisfy the original equation. For this exercise, we got \( y = -2 \) and \( y = 1 \).
Recall that \( y = x^{1/3} \). Hence:
For \( y = -2 \), \( x^{1/3} = -2 \) implies \( x = (-2)^3 = -8 \)
For \( y = 1 \), \( x^{1/3} = 1 \) implies \( x = 1^3 = 1 \).
Next, substitute these solutions into the original equation to check:
For \( x = -8 \), \( (-8)^{2/3} + (-8)^{1/3} - 2 = 4 - 2 - 2 = 0 \)
For \( x = 1 \), \( 1^{2/3} + 1^{1/3} - 2 = 1 + 1 - 2 = 0 \)
Both solutions satisfy the original equation, confirming their correctness.

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