Problem 6
Question
Find all real solutions. $$x^{4}-x^{2}=0$$
Step-by-Step Solution
Verified Answer
The real solutions are \(x = 0\), \(x = 1\), \(x = -1\).
1Step 1: Factor the Equation
The given equation is \(x^4 - x^2 = 0\). You can factor out \(x^2\) from this expression, resulting in \(x^2(x^2 - 1) = 0\).
2Step 2: Set Each Factor to Zero
Now that the equation is factored, we can use the zero-product property. This means we set each factor equal to zero: \(x^2 = 0\) and \(x^2 - 1 = 0\).
3Step 3: Solve for \(x\) from \(x^2 = 0\)
From \(x^2 = 0\), taking the square root of both sides gives \(x = 0\). This is one of the solutions.
4Step 4: Solve for \(x\) from \(x^2 - 1 = 0\)
For \(x^2 - 1 = 0\), add 1 to both sides to get \(x^2 = 1\). Taking the square root of both sides gives \(x = 1\) or \(x = -1\). These are the other solutions.
5Step 5: List All Real Solutions
Now that we have solved both equations, the real solutions to the original problem are \(x = 0\), \(x = 1\), and \(x = -1\).
Key Concepts
Factoring Algebraic ExpressionsZero-Product PropertyFinding Real Solutions
Factoring Algebraic Expressions
Factoring algebraic expressions is a fundamental skill in solving polynomial equations. It involves expressing a polynomial as a product of its factors, making it easier to solve or simplify. Consider the equation \(x^4 - x^2 = 0\). Notice how both terms contain a common factor of \(x^2\). To factor this expression, we extract \(x^2\), which simplifies the equation to \(x^2(x^2 - 1) = 0\).
- Look for common factors in terms.
- Factor out the greatest common factor (GCF).
- Rewrite the equation as a product.
Zero-Product Property
The zero-product property is a handy principle when solving equations. It asserts that if a product of factors is zero, at least one of the factors must be zero. For the factored equation \(x^2(x^2 - 1) = 0\), this property simplifies solving. By setting each factor to zero:
- \(x^2 = 0\)
- \(x^2 - 1 = 0\)
Finding Real Solutions
This step finalizes the process where we determine the actual values of \(x\) that satisfy the equation. After factoring and applying the zero-product property, we solve each separate equation. Take \(x^2 = 0\). Solving gives \(x = 0\) (since the square root of 0 is 0). Moving to \(x^2 - 1 = 0\), we rearrange it to \(x^2 = 1\). Solving this gives us \(x = 1\) or \(x = -1\), because the square root of 1 can be both positive and negative.
Thus, the real solutions to the original polynomial equation \(x^4 - x^2 = 0\) are:
Thus, the real solutions to the original polynomial equation \(x^4 - x^2 = 0\) are:
- \(x = 0\)
- \(x = 1\)
- \(x = -1\)
Other exercises in this chapter
Problem 5
For each quadratic function defined , (a) write the function in the form \(P(x)=a(x-h)^{2}+k,\) (b) give the vertex of the parabola, and (c) graph the function.
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For each complex number, (a) state the real part, (b) state the imaginary part, and (c) identify the number as one or more of the following: real, pure imaginar
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Find a cubic polynomial in standard form with real coefficients, having the given zeros. Let the leading coefficient be \(1 .\) Do not use a calculator. 0 and \
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Each expression. Apply the property \(\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\) if necessary. Do not use a calculator. $$\frac{5 x^{3}+x^{2}}{3 x^{2}}$$
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