Problem 40 - Algebra for College Students

Question

Solve each equation. $$6 x^{4}-31 x^{2}+18=0$$

Step-by-Step Solution

Answer

Solutions: \( x = \pm \frac{3\sqrt{2}}{2} \) and \( x = \pm \frac{\sqrt{6}}{3} \).

Step 1: Identify Substitution

The given equation is a quartic equation in the form of \(6x^4 - 31x^2 + 18 = 0\). To simplify, let's identify a substitution. Let \( y = x^2 \). Then, \( x^4 = y^2 \). This transforms the equation into a quadratic: \( 6y^2 - 31y + 18 = 0 \).

Step 2: Use Quadratic Formula

The equation \(6y^2 - 31y + 18 = 0\) is a standard quadratic equation. We'll use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6, b = -31, \text{ and } c = 18 \).

Step 3: Calculate Discriminant

First, calculate the discriminant \( b^2 - 4ac \). Substituting, we get \((-31)^2 - 4 \times 6 \times 18\). This simplifies to \(961 - 432 = 529\).

Step 4: Solve Quadratic Equation

Substitute the discriminant back into the quadratic formula: \[ y = \frac{-(-31) \pm \sqrt{529}}{2 \times 6} = \frac{31 \pm 23}{12} \].This gives the solutions \( y = \frac{54}{12} = 4.5 \) and \( y = \frac{8}{12} = \frac{2}{3} \).

Step 5: Reverse Substitution

Recall \( y = x^2 \). Hence, we now solve \( x^2 = 4.5 \) and \( x^2 = \frac{2}{3} \). The solutions for \( x \) are:- \( x = \pm \sqrt{4.5} \)- \( x = \pm \sqrt{\frac{2}{3}} \).

Step 6: Simplify Solutions

Calculate the specific values: - \( \sqrt{4.5} = \sqrt{\frac{9}{2}} = \frac{3\sqrt{2}}{2} \), so \( x = \pm \frac{3\sqrt{2}}{2} \).- \( \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \), so \( x = \pm \frac{\sqrt{6}}{3} \). Thus, \( x \) can be \( \pm \frac{3\sqrt{2}}{2} \) or \( \pm \frac{\sqrt{6}}{3} \).

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving equations of the form \( ax^2 + bx + c = 0 \). It provides a straightforward way to find the solutions by plugging coefficients into the formula:

• \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \) is the coefficient of \( y^2 \), \( b \) is the coefficient of \( y \), and \( c \) is the constant term.
This formula is derived from completing the square or the general quadratic equation and is particularly useful when factoring is complex or impossible.
People often use the quadratic formula because:

• It works for every quadratic equation, regardless of whether the roots are real or complex.
• It provides exact answers instead of estimates.

In our exercise, it helped us solve the quadratic formed by the substitution method and yielded solutions for \( y \).

Substitution Method

Substitution is a fantastic technique for simplifying complex equations by replacing variables. It often involves making a temporary substitution for a part of the equation to turn it into a more familiar form.
In the problem at hand, the original equation \( 6x^4 - 31x^2 + 18 = 0 \) is a quartic equation. Solving quartic equations directly can be challenging, so a substitution like \( y = x^2 \) helps.

• By letting \( y = x^2 \), we transform our quartic equation into a simpler quadratic form: \( 6y^2 - 31y + 18 = 0 \).

Once this substitution is made, the problem becomes more straightforward, as solving quadratic equations is typically easier than quartic ones.
After finding solutions for \( y \), we reverse the substitution to express those solutions back in terms of \( x \). This means solving \( x^2 = y \) for each solution of \( y \).

Discriminant Calculation

The discriminant in a quadratic equation looks at the part under the square root in the quadratic formula:

• \( b^2 - 4ac \)

This value is crucial because it tells us a lot about the nature of the roots of the equation.
In the exercise, the discriminant \( (b^2 - 4ac) = 529 \) gave us essential information.

• If the discriminant is positive, like in our case, there are two distinct real roots.
• If it were zero, there would be exactly one real root (a repeating root).
• A negative discriminant would mean the roots are complex and not real.

Calculating the discriminant before solving the quadratic helps anticipate the type of solutions you'll encounter.
In summary, knowing the discriminant allows one to understand not only how many solutions there are but also their nature—real versus complex.