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Magazine Chapter 11: Problem 16
Solve. $$ x^{4}-10 x^{2}+9=0 $$
Short Answer
Expert verified
The solutions are \(x = \pm 1\) and \(x = \pm 3\).
Step by step solution
01
Substitute Variable
Let us start by making a substitution to simplify the problem. Set y = x^{2}. Therefore, we replace every instance of x^{2} in the equation with y.
02
Rewrite the Equation
Substituting y for x^{2}, the equation becomes: \[y^{2} - 10y + 9 = 0\]. Now, we need to solve this quadratic equation.
03
Solve the Quadratic Equation
The quadratic equation \(y^{2} - 10y + 9 = 0\) can be factored. It factors to: \((y - 1)(y - 9) = 0\).
04
Find the Solutions for y
Set each factor equal to zero and solve for y: \[y - 1 = 0\] gives \(y = 1\), and \[y - 9 = 0\] gives \(y = 9\). So, the solutions for y are y = 1 and y = 9.
05
Substitute Back for x
Recall that we set y = x^{2}. Substitute back into this substitution: \[x^{2} = 1\] and \[x^{2} = 9\].
06
Solve for x
Solve the equations \[x^{2} = 1\] and \[x^{2} = 9\]: For \(x^{2} = 1\), x can be \(\pm 1\). For \(x^{2} = 9\), x can be \(\pm 3\).
07
List all Solutions
Combine all the solutions: The solutions to the original equation are \(x = \pm 1\) and \(x = \pm 3\).
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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 faced with complicated equations. It simplifies the problem by introducing a new variable to replace a complex expression. Here鈥檚 how it works in our example:
We have the equation:
\(x^{4}-10 x^{2}+9=0\)
To simplify, we make a substitution. Set \(y = x^{2}\). This reduces the complexity by turning it into a quadratic equation. Essentially, you鈥檙e trading a higher-degree term for a lower-degree one, which is easier to work with.
Here鈥檚 what we do step by step:
Now, any approach we use to solve this new equation will be simpler, thanks to our substitution.
We have the equation:
\(x^{4}-10 x^{2}+9=0\)
To simplify, we make a substitution. Set \(y = x^{2}\). This reduces the complexity by turning it into a quadratic equation. Essentially, you鈥檙e trading a higher-degree term for a lower-degree one, which is easier to work with.
Here鈥檚 what we do step by step:
- Identify a substitution: \(y = x^{2}\)
- Replace all instances of the complex term in the equation: \(y^{2} - 10y + 9 = 0\)
Now, any approach we use to solve this new equation will be simpler, thanks to our substitution.
factoring
Factoring is a method used to solve quadratic equations, making them easier to handle. After applying substitution and rewriting the equation as \(y^{2} - 10y + 9 = 0\), we proceed to factor it.
Factoring involves expressing the equation as a product of its simplest terms. In this case, the quadratic equation factors neatly:
\((y - 1)(y - 9) = 0\)
Each term here is a factor of the quadratic. To find the solutions, we set each factor equal to zero:
Factoring can simplify solutions, especially when terms factor easily. By breaking down the equation into these components, it becomes straightforward to solve.
Factoring involves expressing the equation as a product of its simplest terms. In this case, the quadratic equation factors neatly:
\((y - 1)(y - 9) = 0\)
Each term here is a factor of the quadratic. To find the solutions, we set each factor equal to zero:
- \(y - 1 = 0\) -> \(y = 1\)
- \(y - 9 = 0\) -> \(y = 9\)
Factoring can simplify solutions, especially when terms factor easily. By breaking down the equation into these components, it becomes straightforward to solve.
solving quadratic equations
Solving quadratic equations is a fundamental skill in math. Once substitution and factoring are applied, the solutions to our quadratic equation \(y^{2} - 10y + 9 = 0\) are \(y = 1\) and \(y = 9\).
But remember, we initially substituted \(y\) for \(x^{2}\). So, we need to revert back:
Solving for \(x\), we get:
Finally, we combine these results to list all solutions: \(x = \pm 1\) and \(x = \pm 3\).
We have thus solved the original fourth-degree equation using substitution and factoring, demonstrating the power of breaking down a problem into more manageable parts.
But remember, we initially substituted \(y\) for \(x^{2}\). So, we need to revert back:
- \(x^{2} = 1\)
- \(x^{2} = 9\)
Solving for \(x\), we get:
- From \(x^{2} = 1\), \(x = \pm 1\)
- From \(x^{2} = 9\), \(x = \pm 3\)
Finally, we combine these results to list all solutions: \(x = \pm 1\) and \(x = \pm 3\).
We have thus solved the original fourth-degree equation using substitution and factoring, demonstrating the power of breaking down a problem into more manageable parts.
