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Chapter 3: Problem 6

Find all real solutions. $$x^{4}-x^{2}=0$$

Short Answer

Expert verified
The real solutions are \(x = 0\), \(x = 1\), \(x = -1\).

Step by step solution

01

Factor the Equation

The given equation is \(x^4 - x^2 = 0\). You can factor out \(x^2\) from this expression, resulting in \(x^2(x^2 - 1) = 0\).
02

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 \(x^2 - 1 = 0\).
03

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.
04

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.
05

List All Real Solutions

Now that we have solved both equations, the real solutions to the original problem are \(x = 0\), \(x = 1\), and \(x = -1\).

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Key Concepts

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

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.
Factoring reduces complexity and allows other properties, like the zero-product property, to be applied easily.
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:
  • \(x^2 = 0\)
  • \(x^2 - 1 = 0\)
Using the zero-product property means working with simpler equations. This strategy shortens the solving process by directly leading us to potential solutions.
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:
  • \(x = 0\)
  • \(x = 1\)
  • \(x = -1\)
This method explicitly finds all possible \(x\)-values, ensuring we leave no solution undiscovered.

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Most popular questions from this chapter

For each polynomial function, do the following in order. (a) Use Descartes' rule of signs to find the possible number of positive and negative real zeros. (b) Use the rational zeros theorem to determine the possible rational zeros of the function. (c) Find the rational zeros, if any. (d) Find all other real zeros, if any. (e) Find any other nonreal complex zeros, if any. (f) Find the \(x\) -intercepts of the graph, if any. (g) Find the \(y\) -intercept of the graph. (h) Use synthetic division to find \(P(4),\) and give the coordinates of the corresponding point on the graph. (i) Determine the end behavior of the graph. (i) Sketch the graph. (You may wish to support your answer with a calculator graph.) $$P(x)=3 x^{4}-14 x^{2}-5$$

Use the rational zeros theorem to completely factor \(P(x)\). $$P(x)=6 x^{4}-5 x^{3}-11 x^{2}+10 x-2$$

Use the rational zeros theorem to completely factor \(P(x)\). $$P(x)=5 x^{4}+8 x^{3}-19 x^{2}-24 x+12$$

Divide. $$\frac{3 x^{4}+2 x^{3}-x^{2}+4 x-3}{x^{2}+x-1}$$

For each polynomial, at least one zero is given. Find all others analytically. $$P(x)=x^{4}-41 x^{2}+180 ; \quad-6 \text { and } 6$$
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