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Magazine Chapter 5: Problem 106
Solve \(x^{2 / 3}-4 x^{1 / 3}-5=0\)
Short Answer
Expert verified
\(x = 125\) and \(x = -1\)
Step by step solution
01
- Identify a substitution
Notice that the equation involves terms with exponents involving fractions. Let鈥檚 make a substitution to simplify it. Set \(y = x^{1/3}\). This means that \(y^2 = (x^{1/3})^2 = x^{2/3}\).
02
- Substitute into the equation
Substitute \(y\) and \(y^2\) back into the original equation. Doing this gives us: \(y^2 - 4y - 5 = 0\).
03
- Factor the quadratic equation
Now, we need to solve the quadratic equation \(y^2 - 4y - 5 = 0\). We can factor it as \((y - 5)(y + 1) = 0\).
04
- Solve for \(y\)
Set each factor to zero and solve for \(y\):- \(y - 5 = 0 \rightarrow y = 5\)- \(y + 1 = 0 \rightarrow y = -1\)
05
- Substitute back to find \(x\)
Recall that \(y = x^{1/3}\). Substitute back:- If \(y = 5\), then \(x^{1/3} = 5\rightarrow x = 5^3 = 125\).- If \(y = -1\), then \(x^{1/3} = -1\rightarrow x = (-1)^3 = -1\).
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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 powerful tool for solving equations, especially when they feature terms with fractions or complex exponents. In this problem, we make a substitution to simplify the original equation:
- First, notice that the original equation involves terms like \(x^{2/3}\) and \(x^{1/3}\).
- We can make the substitution \(y = x^{1/3}\) to turn the complicated terms into simpler forms: \(y^2 = x^{2/3}\).
- Substituting these into the equation transforms it into a more familiar quadratic form: \(y^2 - 4y - 5 = 0\).
solving quadratic equations
Solving quadratic equations involves finding the roots of the equation. In this problem, we converted the original equation into a quadratic equation: \[y^2 - 4y - 5 = 0\]
- We solve this quadratic equation by factoring. Factoring means expressing it in the form of \((y - a)(y + b) = 0\).
- For our equation, this means finding two numbers that multiply to -5 and add up to -4. These numbers are 5 and -1.
- \(y - 5 = 0\rightarrow y = 5\)
- \(y + 1 = 0\rightarrow y = -1\)
factoring
Factoring is a technique used to break down more complex expressions into simpler parts that can be multiplied to give the original equation. It's particularly useful for quadratic equations. Here's how it works:
- For the equation \(y^2 - 4y - 5 = 0\), we look for two numbers that multiply to the constant term (-5) and add to the linear coefficient (-4).
- The numbers we need are 5 and -1 because \(5 \times (-1) = -5\) and \(5 + (-1) = 4\).
rational exponents
Rational exponents are expressions where the exponent is a fraction. They represent roots and powers of numbers. In this problem, we have terms with exponents like \(x^{2/3}\) and \(x^{1/3}\):
- \(x^{1/3}\) represents the cube root of x, and \(x^{2/3}\) represents the square of the cube root of x.
- Understanding these allows us to make a useful substitution \((y = x^{1/3})\) to simplify the equation.
- For \(y = 5\), \(x^{1/3} = 5\rightarrow x = 5^3 = 125\).
- For \(y = -1\), \(x^{1/3} = -1\rightarrow x = (-1)^3 = -1\).
