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Magazine Chapter 2: Problem 18
Exer. 1-50: Solve the equation. $$ \sqrt[4]{2 x^{2}-1}=x $$
Short Answer
Expert verified
x = 1 is the solution.
Step by step solution
01
Eliminate the Fourth Root
To remove the fourth root, raise both sides of the equation to the power of 4. This gives us: \[ (\sqrt[4]{2x^2 - 1})^4 = x^4 \]which simplifies to:\[ 2x^2 - 1 = x^4 \]
02
Rearrange the Equation
Bring all terms to one side of the equation to set it equal to zero:\[ x^4 - 2x^2 + 1 = 0 \]
03
Substitute to Simplify
Let \( y = x^2 \). Then \( y^2 = x^4 \). Substituting into the equation gives:\[ y^2 - 2y + 1 = 0 \]
04
Solve the Quadratic Equation
Notice that the quadratic equation is a perfect square trinomial:\[ (y - 1)^2 = 0 \]Therefore, solve for \( y \):\[ y - 1 = 0 \ y = 1 \]
05
Back-Substitute for x
Since \( y = x^2 \), substitute back to find \( x \):\[ x^2 = 1 \]Solving for \( x \) gives two solutions:\[ x = 1 \quad \text{or} \quad x = -1 \]
06
Verify the Solutions
Check both solutions in the original equation:1. For \( x = 1 \): \[ \sqrt[4]{2(1)^2 - 1} = \sqrt[4]{1} = 1 \] which matches2. For \( x = -1 \): \[ \sqrt[4]{2(-1)^2 - 1} = \sqrt[4]{1} = 1 \] which does not match because \( -1 eq 1 \)Thus, the only valid solution is \( x = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Fourth Root
The fourth root of a number is simply the number that, when multiplied by itself four times, returns the original number. For example, the fourth root of 16 is 2, because \(2 \times 2 \times 2 \times 2 = 16\). To calculate the fourth root, you can use the notation \( \sqrt[4]{x} \), where \( x \) is the given number.
- The fourth root can be removed by raising the expression to the 4th power.
- In the equation \( \sqrt[4]{2x^2 - 1} = x \), applying this principle helps to simplify the equation by removing the root.
Solving the Quadratic Equation
Once the equation \( (\sqrt[4]{2x^2 - 1})^4 = x^4 \) was simplified to \( x^4 - 2x^2 + 1 = 0 \), it became a polynomial equation in a fourth degree.
- By substituting \( y = x^2 \), the problem was transformed into a quadratic equation: \( y^2 - 2y + 1 = 0 \).
- This simplification is essential because quadratic equations are easier to handle and solve compared to quartic ones.
Recognizing a Perfect Square Trinomial
The quadratic equation \( y^2 - 2y + 1 = 0 \) in this exercise is a perfect square trinomial.
- A perfect square trinomial is a type of quadratic expression that can be written as the square of a binomial.
- It takes the form \( (a - b)^2 = a^2 - 2ab + b^2 \).
Verification of Solutions
Verification is a crucial step in solving equations that helps ensure the solutions are correct. After determining the possible solutions, \( x = 1 \) and \( x = -1 \), it is important to plug these values back into the original equation \( \sqrt[4]{2x^2 - 1} = x \).
- For \( x = 1 \), the equation holds true: \( \sqrt[4]{2(1)^2 - 1} = 1 \).
- For \( x = -1 \), the equation does not hold: \( \sqrt[4]{2(-1)^2 - 1} = 1 \), but we end up with \( -1 eq 1 \).
