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Solve the given problems. All numbers are accurate to at least two significant digits. Solve the equation \(x^{4}-5 x^{2}+4=0\) for \(x\). [Hint: The equation can be written as \(\left.\left(x^{2}\right)^{2}-5\left(x^{2}\right)+4=0 . \text { First solve for } x^{2} .\right]\)

Short Answer

Expert verified
The solutions are \( x = 2, -2, 1, \text{ and } -1 \).

Step by step solution

01

Identify the substitution

Notice that the given equation is a quadratic in form if we let \( y = x^2 \). By substituting \( y \) for \( x^2 \), the equation \( x^4 - 5x^2 + 4 = 0 \) transforms to \( y^2 - 5y + 4 = 0 \).
02

Solve the quadratic equation

Now, solve the quadratic equation \( y^2 - 5y + 4 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -5 \), and \( c = 4 \).
03

Apply the quadratic formula

Plug the values into the quadratic formula: \[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2} \].
04

Find the solutions for y

Calculate: \( y = \frac{5 + 3}{2} = 4 \) and \( y = \frac{5 - 3}{2} = 1 \). Thus, we have two possible values for \( y \): 4 and 1.
05

Substitute back for x

Recall that \( y = x^2 \). Therefore, solve \( x^2 = 4 \) and \( x^2 = 1 \).
06

Solve for x

For \( x^2 = 4 \), the solutions are \( x = \pm 2 \). For \( x^2 = 1 \), the solutions are \( x = \pm 1 \).
07

Compile all solutions

Thus, the solutions for \( x \) are \( x = 2, -2, 1, \text{ and } -1 \).

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Key Concepts

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

Polynomial Equations
Polynomial equations are expressions that set polynomials equal to zero. They may contain variables with exponents, constants, and the operations of addition, subtraction, and multiplication. Understanding these forms is fundamental in algebra. The highest power of the variable in a polynomial equation determines its degree.
In the example from the exercise, we dealt with a quartic equation: \( x^4 - 5x^2 + 4 = 0 \), which is a fourth-degree polynomial. These problems can often be complex, so simplifying them using substitutions or factoring is common.
It's important to recognize the patterns in polynomials, such as when they can be perceived as a quadratic form by substituting. This recognition can simplify the solution process immensely. When encountering polynomial equations in math, always check for opportunities to reduce their complexity through these techniques.
Quadratic Formula
The quadratic formula is a powerful tool to solve quadratic equations. It provides the solutions to any equation of the form \( ax^2 + bx + c = 0 \), by finding values of \( x \) that make the equation true.
To use the quadratic formula:
  • Identify the coefficients \( a \), \( b \), and \( c \) from the equation.
  • Substitute these into: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our exercise, after substituting \( y = x^2 \), the equation became \( y^2 - 5y + 4 = 0 \). Using \( a = 1 \), \( b = -5 \), and \( c = 4 \), we were able to determine the solutions for \( y \).
This systematic approach reveals possible roots by processing each calculation step-by-step. The roots calculated from the quadratic formula may need further steps in substitution exercises, as we observed when going back to solve for \( x \) from \( y = x^2 \). The quadratic formula not only simplifies finding solutions but also is applicable to a vast range of algebraic problems.
Algebraic Substitution
Algebraic substitution is a technique used to simplify complex equations by replacing parts of the equation with a new variable. This method is a strategic move that transforms the equation into a more recognizable form.
In our textbook example, the given polynomial was first rewritten by substituting \( y = x^2 \). This replaced the original equation \( x^4 - 5x^2 + 4 = 0 \) with a simpler quadratic form: \( y^2 - 5y + 4 = 0 \).
The substitution served as a bridge to apply the quadratic formula effectively. After solving for \( y \), the substitution was reversed by solving \( x^2 = y \) for each obtained value of \( y \).
This process is typical in algebra as it tackles high-degree polynomials by breaking them into multiple manageable parts. It's a versatile approach, allowing for greater flexibility in problem-solving.

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