Problem 34 - Algebra for College Students

Question

Solve each equation. $$x^{4}-21 x^{2}+54=0$$

Step-by-Step Solution

Answer

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

Step 1: Substitute Variable

To simplify solving the equation, substitute a new variable \(y = x^2\). Then rewrite the original equation in terms of \(y\): \(y^2 - 21y + 54 = 0\).

Step 2: Identify Equation Type

The equation \(y^2 - 21y + 54 = 0\) is a quadratic equation, so we will use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=1\), \(b=-21\), and \(c=54\).

Step 3: Calculate Discriminant

Calculate the discriminant (\(D\)) using the formula \(D = b^2 - 4ac\). Substitute \(a = 1\), \(b = -21\), and \(c = 54\) into the formula: \[D = (-21)^2 - 4 \times 1 \times 54 = 441 - 216 = 225\].

Step 4: Solve Quadratic Equation

With \(D = 225\), apply the quadratic formula to find \(y\): \[y = \frac{-(-21) \pm \sqrt{225}}{2 \times 1} = \frac{21 \pm 15}{2}\]. Calculate the two possible values of \(y\): \[y_1 = \frac{21 + 15}{2} = 18\] and \[y_2 = \frac{21 - 15}{2} = 3\].

Step 5: Re-substitute Original Variable

Recall that \(y = x^2\). Substitute back to find \(x\): for \(y = 18\), \(x^2 = 18\) leading to \(x = \pm \sqrt{18}\) or \(x = \pm 3\sqrt{2}\). For \(y = 3\), \(x^2 = 3\) leading to \(x = \pm \sqrt{3}\).

Step 6: Compile Solution

Gather all solutions found: \(x = \pm 3\sqrt{2}\) and \(x = \pm \sqrt{3}\). There are four solutions to the original equation.

Key Concepts

Substitution Method

The substitution method is a clever technique used to simplify complex equations, often transforming them into a more manageable form. In our exercise, the equation involves a fourth-degree polynomial. Directly solving fourth-degree equations can be quite challenging, hence the need for substitution.

We start by introducing a new variable, substituting a simpler equivalent to a part of the equation. In our case, we let the new variable be defined as follows:

• Substitute: Let \( y = x^2 \).

This transforms the original equation \(x^4 - 21x^2 + 54 = 0\) into the quadratic equation \(y^2 - 21y + 54 = 0\). This substitution significantly simplifies our task since quadratic equations are much easier to solve compared to fourth-degree polynomials. This method is particularly useful when the original equation can be expressed in terms of perfect squares, allowing a straightforward translation into a one-variable quadratic format.

Solving Polynomial Equations

Polynomial equations, like the one in the example, contain variables raised to positive integer powers. Here, we're presented with a polynomial of degree four initially. Solving these equations generally involves finding the roots, or the values of the variable that make the polynomial equation equal to zero.

After the substitution step, our polynomial in terms of \( y \) becomes quadratic, simplifying the task. With a quadratic polynomial \( y^2 - 21y + 54 = 0 \), the typical methods to solve include factoring, completing the square, or using the quadratic formula. Using

• Substitution first. This transforms the equation into an easier quadratic form.
• Employing the Quadratic Formula or other algebraic techniques to find solutions.

Through these methods, we extract the possible values for \( y \), which then contribute to solving for the substituted variable \( x \). It showcases how breaking down complex polynomial structures using substitutions and known algebraic solutions can efficiently lead to discovering roots of higher-order polynomials.

Quadratic Formula

The quadratic formula is a universal tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It allows us to find the roots of the equation without requiring factoring. The formula is expressed as:

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

In our context, the equation \( y^2 - 21y + 54 = 0 \) fits this format perfectly with \( a = 1 \), \( b = -21 \), and \( c = 54 \).

After calculating the discriminant, we used this formula to find:

• The distinct roots, \(y_1 = 18\) and \(y_2 = 3\), for our quadratic equation.

This solution path highlights the quadratic formula’s versatility in tackling any quadratic equation, irrespective of its complexity. Utilizing the quadratic formula ensures that all solutions are exhausted, and it's notably effective when factoring is not straightforward.

Discriminant

The discriminant is a crucial component of the quadratic formula, represented by the expression \( D = b^2 - 4ac \). It provides valuable information about the nature and number of roots of a quadratic equation:

• If \( D > 0 \), there are two distinct real roots.
• If \( D = 0 \), there is exactly one real root, often implying a double root.
• If \( D < 0 \), the roots are complex and not real.

In our exercise, the calculated discriminant \( D = 225 \) is a positive perfect square, indicating two distinct real roots. This confirmed the applicability of the quadratic formula in solving for \( y \).

Understanding the discriminant is an essential step in predicting the outcome of a quadratic equation even before solving. It gives us a quick understanding of whether we should expect real or complex solutions, guiding our choice of method.