Problem 33 - Algebra for College Students

Question

Solve each equation. $$x^{4}-18 x^{2}+72=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \pm 2\sqrt{3}\) and \(x = \pm \sqrt{6}\).

Step 1: Identify the Quadratic Form

The given equation is \(x^4 - 18x^2 + 72 = 0\), which is a polynomial equation of degree 4. We notice that it can be treated as a quadratic equation in disguise by letting \(y = x^2\). This transforms the equation to a simpler quadratic equation \(y^2 - 18y + 72 = 0\).

Step 2: Solve the Quadratic Equation

The quadratic equation \(y^2 - 18y + 72 = 0\) can be solved using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -18\), and \(c = 72\). Substitute these values into the formula to get:\[y = \frac{-(-18) \pm \sqrt{(-18)^2 - 4 \times 1 \times 72}}{2 \times 1}\] Simplify this to find the values of \(y\).

Step 3: Calculate the Discriminant

First, calculate the discriminant: \((-18)^2 - 4 \times 1 \times 72 = 324 - 288 = 36\). Since the discriminant is positive, we expect two distinct real solutions for \(y\).

Step 4: Find the Roots of the Quadratic

Now, substitute the discriminant back into the quadratic formula:\[y = \frac{18 \pm \sqrt{36}}{2} = \frac{18 \pm 6}{2}\].This gives us:\(y = \frac{24}{2} = 12\) and \(y = \frac{12}{2} = 6\). Thus, the solutions for \(y\) are \(y = 12\) and \(y = 6\).

Step 5: Back-Substitute to Solve for \(x\)

Since \(y = x^2\), replace \(y\) with \(x^2\) in our solutions to find \(x\). For \(x^2 = 12\), take the square root to get \(x = \pm \sqrt{12}\), which simplifies to \(x = \pm 2\sqrt{3}\). For \(x^2 = 6\), take the square root to get \(x = \pm \sqrt{6}\). Thus, the solutions for \(x\) are \(x = \pm 2\sqrt{3}\) and \(x = \pm \sqrt{6}\).

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions that involve variables raised to whole number powers and constants. A polynomial equation can have one or more terms, but the power of the highest term determines its degree. For example, in the given problem, we have an equation \(x^4 - 18x^2 + 72 = 0\), which is a fourth-degree polynomial because the highest power is 4.

• To solve a polynomial equation, we usually seek to find the values of the variable (like \(x\)) that make the equation true, called the roots or solutions.
• In some cases, like our example, a higher degree polynomial might be expressible as a quadratic in a different variable, simplifying the process.

This strategy involves substituting a new variable to transform the polynomial into a simpler quadratic form, as we attempted with \(y = x^2\). The resulting equation, \(y^2 - 18y + 72 = 0\), is much easier to handle with known quadratic solution methods.

Discriminant

The discriminant is a key concept in solving quadratic equations as it tells us how many real roots the equation has. The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the expression \(b^2 - 4ac\).

• If \(\Delta > 0\), the quadratic equation has two distinct real roots.
• If \(\Delta = 0\), there is exactly one real root (a double root).
• If \(\Delta < 0\), there are no real roots; instead, the roots are complex numbers.

In our example, the transformed equation is \(y^2 - 18y + 72 = 0\) with \(a = 1\), \(b = -18\), and \(c = 72\). The discriminant is calculated as \((-18)^2 - 4 \times 1 \times 72 = 36\).
Since \(\Delta = 36\) is positive, it indicates that the equation has two distinct real solutions for \(y\).

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions of a quadratic equation \(ax^2 + bx + c = 0\). It is derived from the process of completing the square and gives the roots of the quadratic as follows:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula is particularly useful because it works for all forms of quadratic equations, regardless of whether the quadratic can be factored easily or not.

• In the expression, \(b \pm \sqrt{b^2 - 4ac}\) implies that there are potentially two solutions due to the \(\pm\) symbol.
• Calculating the square root of the discriminant \(b^2 - 4ac\) gives insight into the nature and number of solutions.

For the equation \(y^2 - 18y + 72 = 0\), we substitute \(a = 1\), \(b = -18\), and \(c = 72\) into the quadratic formula. The solutions are:\[y = \frac{18 \pm \sqrt{36}}{2}\]which simplify to \(y = 12\) and \(y = 6\).
These are then related back to the original variable \(x\) as \(x = \pm 2\sqrt{3}\) and \(x = \pm \sqrt{6}\), completing the solution process for our polynomial equation.