Chapter 0: Problem 4
Solve the following equations: $$ x^{4}-x^{2}=6 $$
Short Answer
Expert verified
The solutions to the equation \(x^4 - x^2 = 6\) are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
Step by step solution
01
Perform substitution to create a quadratic equation
Let's set \(u=x^2\), then our equation becomes:
\(u^2 - u = 6\)
02
Rearrange the quadratic equation
Rearrange the equation to standard quadratic form, which is \(ax^2 + bx + c = 0\). In our case, it would be:
\(u^2 - u - 6 = 0\)
03
Factor the quadratic equation
Now, let's factor our quadratic equation. We are looking for two numbers that multiply to -6 and add up to -1. The numbers we are looking for are -3 and 2. So, our factored equation is:
\((u - 3)(u + 2) = 0\)
04
Solve for u
Now, solve for u by setting each factor equal to zero:
1) \(u - 3 = 0\) => \(u = 3\)
2) \(u + 2 = 0\) => \(u = -2\)
Now we have two solutions for u: 3 and -2.
05
Substitute back x and solve
Now, let's substitute back x into our solutions for u.
For \(u = 3\):
\(x^2 = 3\) => \(x = \pm\sqrt{3}\)
For \(u = -2\):
\(x^2 = -2\)
Since there are no real solutions for the square root of a negative number, there are no solutions for x in this case.
06
Write down the final solutions
So, the solutions to the original equation \(x^4 - x^2 = 6\) are:
\(x = \sqrt{3}\) and \(x = -\sqrt{3}\)
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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 useful technique when dealing with complex equations. In this method, you replace a variable to transform the equation into a simpler form. This approach is particularly effective when handling higher degree polynomials.
For our exercise, we start with the equation \(x^{4} - x^{2} = 6\). To simplify, we let \(u = x^2\). This substitution changes our equation into \(u^2 - u = 6\), simplifying a quartic equation into a quadratic one.
This makes it easier to handle and solve using quadratic solving techniques. Remember, the goal of substitution is to rewrite the equation in a way that is easier to solve.
For our exercise, we start with the equation \(x^{4} - x^{2} = 6\). To simplify, we let \(u = x^2\). This substitution changes our equation into \(u^2 - u = 6\), simplifying a quartic equation into a quadratic one.
This makes it easier to handle and solve using quadratic solving techniques. Remember, the goal of substitution is to rewrite the equation in a way that is easier to solve.
Factoring
Factoring is an essential skill in algebra, especially when solving quadratic equations. The process involves writing a polynomial as a product of its factors. For the quadratic equation \(u^2 - u - 6 = 0\), we need to find two numbers that multiply to \(-6\) and add up to \(-1\).
These numbers are \(-3\) and \(2\), and they help us to rewrite the equation as \((u - 3)(u + 2) = 0\).
Once the equation is factored, it becomes much easier to solve. Each factor represents a potential solution for the equation.
These numbers are \(-3\) and \(2\), and they help us to rewrite the equation as \((u - 3)(u + 2) = 0\).
Once the equation is factored, it becomes much easier to solve. Each factor represents a potential solution for the equation.
- The process involves finding pairs of numbers that satisfy the conditions given by the original equation.
- This technique is not only used in math classes but also in real-world applications.
Solving Equations
Solving equations involves finding the values of variables that satisfy the given mathematical statement. After factoring our quadratic equation into \((u - 3)(u + 2) = 0\), we can find the solutions by setting each factor equal to zero.
This can sometimes involve substituting back to the original terms or working through complex number solutions if necessary.
- First, solve \(u - 3 = 0\) which gives \(u = 3\).
- Then, solve \(u + 2 = 0\) which gives \(u = -2\).
This can sometimes involve substituting back to the original terms or working through complex number solutions if necessary.
Real Solutions
A real solution to an equation is a number that exists within the set of real numbers. In our context, after substitution and solving the equation \(x^2 = u\), it's crucial to identify which solutions remain real.
For \(u = 3\), we find \(x^2 = 3\), which results in real solutions \(x = \pm\sqrt{3}\). These represent the points where our original equation is satisfied in real-number terms.
However, for \(u = -2\), we encounter \(x^2 = -2\), where no real number \(x\) can satisfy this equation since the square root of a negative number is non-real or imaginary.
For \(u = 3\), we find \(x^2 = 3\), which results in real solutions \(x = \pm\sqrt{3}\). These represent the points where our original equation is satisfied in real-number terms.
However, for \(u = -2\), we encounter \(x^2 = -2\), where no real number \(x\) can satisfy this equation since the square root of a negative number is non-real or imaginary.
- Real solutions mean the solution values are part of the realm of real numbers—numbers you can plot on a number line.
- In problems like this, distinguishing real from non-real (complex) solutions is important.