Problem 39 - Algebra for College Students

Question

Solve each equation. $$6 x^{4}-29 x^{2}+28=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \pm \sqrt{3.5}\) and \(x = \pm \frac{2\sqrt{3}}{3}\).

Step 1: Introduce a Substitution

Notice that the equation is in the form of a quadratic in terms of \(x^2\). Let \(u = x^2\), which transforms the equation to \(6u^2 - 29u + 28 = 0\).

Step 2: Quadratic Formula Application

Recognize \(6u^2 - 29u + 28 = 0\) as a quadratic equation in \(u\). Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 6\), \(b = -29\), and \(c = 28\).

Step 3: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\) which is \((-29)^2 - 4 \times 6 \times 28 = 841 - 672 = 169\).

Step 4: Solve for u Using Quadratic Formula

Substitute the discriminant and the coefficients into the quadratic formula: \[ u = \frac{29 \pm \sqrt{169}}{12} \]. Since \(\sqrt{169} = 13\), this simplifies to \[ u = \frac{29 \pm 13}{12} \].

Step 5: Simplification and Solutions for u

Calculate the two different solutions for \(u\): 1. \(u = \frac{29 + 13}{12} = \frac{42}{12} = 3.5\)2. \(u = \frac{29 - 13}{12} = \frac{16}{12} = \frac{4}{3}\)

Step 6: Back Substitute to Find x

Recall that \(u = x^2\), thus, for each \(u\):1. If \(u = 3.5\), then \(x^2 = 3.5\). Taking square roots gives \(x = \pm \sqrt{3.5}\).2. If \(u = \frac{4}{3}\), then \(x^2 = \frac{4}{3}\). Taking square roots gives \(x = \pm \frac{2}{\sqrt{3}}\) or \(x = \pm \frac{2\sqrt{3}}{3}\) after rationalizing the denominator.

Step 7: Present the Final Solutions

The solutions to the original equation \(6x^4 - 29x^2 + 28 = 0\) are:\(x = \pm \sqrt{3.5}\) and \(x = \pm \frac{2\sqrt{3}}{3}\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is 2. These equations are generally in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
For example, if you have an equation such as \( x^2 - 4x + 3 = 0 \), this is a quadratic equation. The equation can have two solutions which can be real or complex numbers.

In some problems, you might encounter equations that can be converted into a quadratic form by substituting a temporary variable. This makes it easier to apply various methods to find the roots or solutions of the original equation.

Substitution Method

The substitution method is a helpful tool when dealing with complex equations, such as higher-degree polynomials that can be reduced to quadratics.

For equations like \( 6x^4 - 29x^2 + 28 = 0 \), which involve higher powers, substitution can simplify the process.

• Start by identifying a suitable substitution. In this case, let \( u = x^2 \).
• This transforms the original equation to a simpler form: \( 6u^2 - 29u + 28 = 0 \).
• Now, you can focus on solving the quadratic equation in terms of \( u \).

By substituting back after solving, you can find the original variable values. This method is especially useful for equations in an "even" polynomial format.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations given by:\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula provides the solutions for any quadratic equation \( au^2 + bu + c = 0 \).

The formula involves:

• The coefficients \( a \), \( b \), and \( c \).
• Calculating the discriminant \( b^2 - 4ac \), which determines the nature of the roots.

For the equation \( 6u^2 - 29u + 28 = 0 \), we have \( a = 6 \), \( b = -29 \), and \( c = 28 \). By inserting these values into the quadratic formula, we can obtain the two possible solutions for \( u \).

Using the formula ensures that we account for all potential roots, regardless of their nature or complexity.

Discriminant Calculation

The discriminant is a part of the quadratic formula and plays a crucial role in determining the nature of the roots of a quadratic equation. It is given by:\[ b^2 - 4ac \]
Depending on the value of the discriminant, the roots can be:

• Positive: Two distinct real roots.
• Zero: Exactly one real root (a repeated root).
• Negative: Two complex roots.

In our example, with \( b = -29 \), \( a = 6 \), and \( c = 28 \), we calculate the discriminant as:\[ (-29)^2 - 4 \times 6 \times 28 = 841 - 672 = 169 \]
Since the discriminant is positive, we expect two distinct real solutions for \( u \). Understanding the discriminant helps predict and verify the type of solutions before applying the quadratic formula.