Problem 39 - College Algebra

Question

Solve each equation in by making an appropriate substitution. $$ x^{4}-5 x^{2}+4=0 $$

Step-by-Step Solution

Answer

The solutions to the equation \(x^{4}-5 x^{2}+4=0\) are \(x = -2, -1, 1, 2\).

Step 1: Substitution

Let's make the substitution \(u = x^{2}\). The equation thus changes to a quadratic equation: \(u^{2} - 5u + 4 = 0\).

Step 2: Solve the quadratic equation

Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula. The given quadratic factors easily into: \((u-4)(u-1) = 0\).

Step 3: Find the roots

Setting the factors to zero gives the solutions for \(u\): \(u=4\) and \(u=1\).

Step 4: Back substitution

Substitute \(u\) back in terms of \(x^{2}\) and solve: \(x^{2} = 4\) gives \(x = \pm 2\), and \(x^{2} = 1\) gives \(x = \pm 1\).

Key Concepts

Substitution Method

The substitution method is a clever technique used to simplify complex equations, transforming them into a form that is easier to work with. Here's how it works:

• Identify a part of the equation that can be substituted with a simpler expression. In our original exercise, we substitute the square term, making the equation simpler to handle.
• Replace the identified term with a single variable. For instance, in this problem, we let \( u = x^2 \). This changes the original quartic equation, \( x^4 - 5x^2 + 4 = 0 \), into the quadratic equation \( u^2 - 5u + 4 = 0 \).

This method is highly effective when you have polynomial equations like these, where a substitution can transform it into a common quadratic form. Soon, it'll be ready for methods like factoring or using the quadratic formula to find solutions. After solving for the new variable \( u \), you'll substitute back to find the original variable, \( x \). This step-by-step transformation makes complex problems much more manageable.

Factoring Quadratics

Factoring is a popular method to solve quadratic equations when they can be easily broken down into simpler binomial expressions. Here's how factoring works:

• Look for two numbers that multiply to give the constant term (the third term in the quadratic expression) and add to give the coefficient of the linear term (the second term).
• For \( u^2 - 5u + 4 = 0 \), the numbers \(-1\) and \(-4\) fulfill these conditions because \(-1 \times -4 = 4\) and \(-1 + -4 = -5\).
• Split the quadratic expression into the product of two binomials: \((u - 4)(u - 1) = 0\).

Factoring allows for quick solutions as it splits the equation into products that equal zero. Each factor can then be set to zero, giving direct solutions for \( u \). The solutions \( u = 4 \) and \( u = 1 \) are then used for back substitution into the original variable, allowing us to find \( x \). This method is quick and efficient for quadratics that can be easily factored. However, not all quadratics will factor so simply, and that's when the quadratic formula comes into play.

Quadratic Formula

The quadratic formula is a universal tool for solving any quadratic equation, whether they factor easily or not. It is particularly useful when the expression does not factor neatly. The formula is:a\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's a simple guide on how to use it:

• Identify the coefficients \( a \), \( b \), and \( c \) from the quadratic equation in the form \( ax^2 + bx + c = 0 \).
• Substitute these values into the quadratic formula. For \( u^2 - 5u + 4 = 0 \), \( a = 1 \), \( b = -5 \), and \( c = 4 \).
• Calculate the discriminant \( b^2 - 4ac \) to ensure it is not negative. A positive discriminant gives real roots, while a zero discriminant gives one real root.
• Use the formula to find the roots of the equation. Plugging in our values, we obtain the same roots \( u = 4 \) and \( u = 1 \) as derived using factoring.

The quadratic formula can always provide a solution when others can't. It acts as a foolproof method, especially in instances where other techniques, like factoring, hit a roadblock. Understanding it fully not only helps in solving equations but also builds a strong mathematical foundation.