Problem 16 - Precalculus: Functions and Graphs

Question

Solve the equation by using the special quadratic equation on page 53. \(16 x^{2}=49\)

Step-by-Step Solution

Answer

The solutions are \(x = \frac{7}{4}\) and \(x = -\frac{7}{4}\).

Step 1: Identify the Structure of the Equation

The given equation is \(16x^2 = 49\). Notice that this equation is already structured as a quadratic equation equal to another constant: \(ax^2 = c\). In this case, \(a = 16\) and \(c = 49\), which allows us to use special strategies for solving equations in this form.

Step 2: Isolate the Quadratic Term

To isolate \(x^2\), divide both sides of the equation by 16:\[x^2 = \frac{49}{16}\] This isolates the quadratic term and transforms the equation into a simpler form.

Step 3: Apply the Square Root

To solve for \(x\), take the square root of both sides of the equation:\[x = \pm\sqrt{\frac{49}{16}}\]This step means we consider both positive and negative roots since we are breaking open the power of 2.

Step 4: Simplify the Expression

Simplify the square root expression:\[x = \pm \frac{\sqrt{49}}{\sqrt{16}}\]Since \(\sqrt{49} = 7\) and \(\sqrt{16} = 4\), the equation becomes:\[x = \pm \frac{7}{4}\]

Step 5: State the Solution

The solutions are \(x = \frac{7}{4}\) and \(x = -\frac{7}{4}\). These are the roots of the original equation and satisfy the condition \(16x^2 = 49\).

Key Concepts

Solving Quadratic Equations

Quadratic equations are fundamental algebraic expressions of the form \(ax^2 + bx + c = 0\). These equations can appear daunting at first. However, solving them involves a systematic approach to discovering the value of \(x\) that satisfies the equation. The equation \(16x^2 = 49\) is a quadratic equation with a structure of \(ax^2 = c\). Here, there is no linear term, meaning it is set up perfectly for direct solving. To solve such equations, we generally aim to express \(x^2\) in terms of constants, enabling the application of methods like taking square roots or using the quadratic formula. Knowing how to identify the type of quadratic equation helps in selecting the most efficient method to find its solutions.

Square Root Method

The square root method is a straightforward way to solve quadratic equations that are already arranged in a specific form, such as \(ax^2 = c\). For the given equation \(16x^2 = 49\), we start by isolating \(x^2\). This is achieved by dividing both sides by the coefficient of \(x^2\), which is 16 in this case:

• Equation: \(16x^2 = 49\)
• Divide both sides by 16: \(x^2 = \frac{49}{16}\)

Once \(x^2\) is isolated, the next step is to take the square root of both sides:

• Apply square root: \(x = \pm\sqrt{\frac{49}{16}}\)

This yields two possible values for \(x\) because squaring both a positive and negative number gives the same result. To finalize, simplify the square root:

• Simplified square root: \(x = \pm\frac{7}{4}\)

The square root method is efficient for quadratic equations lacking the linear \(bx\) component. It provides quick solutions with minimal steps, making it an excellent choice when applicable.

Quadratic Formula

When a quadratic equation is not neatly set up like \(ax^2 = c\), or when coefficients make other methods difficult, the quadratic formula provides a reliable method for finding solutions. For a standard quadratic equation \(ax^2 + bx + c = 0\), the formula is:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula derives solutions by considering all possible configurations of the quadratic expression, including those cases where the roots may not be rational. The part under the square root, known as the discriminant \(b^2 - 4ac\), tells us the nature of the roots:

• If \(b^2 - 4ac > 0\), there are two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root (repeated).
• If \(b^2 - 4ac < 0\), there are no real roots (the roots are complex).

The quadratic formula is incredibly useful because it applies to any quadratic equation, regardless of its specific structure or coefficients. It's a versatile tool in the algebra toolkit.