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Magazine Chapter 5: Problem 10
In \(3-18,\) find all roots of each given function by factoring or by using the quadratic formula. $$ f(x)=x^{4}+5 x^{2}+4 $$
Short Answer
Expert verified
The roots are \( x = i, -i, 2i, \) and \(-2i\).
Step by step solution
01
Recognize the Substitution
Notice that the function \( f(x) = x^4 + 5x^2 + 4 \) looks similar to a quadratic form. Let \( y = x^2 \). Then, the function can be rewritten as \( y^2 + 5y + 4 \).
02
Factor the Quadratic Expression
The expression \( y^2 + 5y + 4 \) can now be factored. We look for two numbers that multiply to 4 and add to 5. These numbers are 1 and 4. Therefore, \( y^2 + 5y + 4 \) factors to \((y + 1)(y + 4)\).
03
Substitute Back to Original Variable
Since \( y = x^2 \), substitute back the value of \( y \) into the factors: \((x^2 + 1)(x^2 + 4)\).
04
Find the Roots from Factored Form
Set each factor equal to zero to find the roots of \( f(x) \): \( x^2 + 1 = 0 \) and \( x^2 + 4 = 0 \).
05
Solve for Roots of Quadratic Equations
For \( x^2 + 1 = 0 \), subtract 1 from both sides to get \( x^2 = -1 \). The solutions are \( x = i \) and \( x = -i \), where \( i \) is the imaginary unit. For \( x^2 + 4 = 0 \), subtract 4 from both sides to get \( x^2 = -4 \). The solutions here are \( x = 2i \) and \( x = -2i \).
06
Compile all Roots
The roots of the original function \( f(x) = x^4 + 5x^2 + 4 \) are \( x = i, -i, 2i, \) and \(-2i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). It comes in very handy especially when factoring is not straightforward. The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula gives us the roots of the quadratic equation by considering the coefficients \(a\), \(b\), and \(c\). Here's how it works:
- Calculate the discriminant \(D = b^2 - 4ac\). This determines the number and type of roots:
- If \(D > 0\), there are two distinct real roots.
- If \(D = 0\), there is exactly one real root (a repeated root).
- If \(D < 0\), the roots are complex and appear as complex conjugates.
- Substitute \(a\), \(b\), and \(c\) into the formula to find \(x\).
Imaginary Numbers
Imaginary numbers arise when we take square roots of negative numbers. For example, the square root of -1 is represented by the symbol \(i\), which is called the imaginary unit. It's essential to note:
- An imaginary number is of the form \(bi\) where \(b\) is a real number and \(i\) stands for \(\sqrt{-1}\).
- A complex number includes both a real part and an imaginary part, expressed as \(a + bi\).
- \(x^2 = -1\) leads to \(x = i\) and \(x = -i\).
- Similarly, \(x^2 = -4\) gives solutions \(x = 2i\) and \(x = -2i\), since \(\sqrt{-4} = \pm2i\).
Solving Quadratic Equations
Solving quadratic equations is a fundamental skill in algebra. Every quadratic equation can be solved by a few different methods, and choosing the appropriate one depends on the particular equation. Here are the common approaches:
- **Factoring**: This involves expressing the quadratic equation as a product of two binomials. For example, the equation \(x^2 + 5x + 4\) can be factored as \((x + 1)(x + 4)\).
- **Quadratic Formula**: As discussed, this provides a direct way to find the roots using the coefficients of the equation.
- **Completing the Square**: Another method which involves rewriting the equation in the form \((x + p)^2 = q\), and solving for \(x\).
