Problem 83 - Algebra for College Students

Question

Find all real or imaginary solutions to each equation. Use the method of your choice. $$\left(p+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Step-by-Step Solution

Answer

The solutions are \(p = 1\) and \(p = -2\).

Step 1: Rewrite the Equation

Start by recognizing that the equation \(\big(p+\frac{1}{2}\big)^{2}=\frac{9}{4}\) is a quadratic equation in the form of a square. This can help identify potential solutions by taking the square root on both sides.

Step 2: Take the Square Root of Both Sides

To isolate \(p + \frac{1}{2}\), take the square root of both sides: \(\big(p+\frac{1}{2}\big)= \pm\frac{3}{2}\). This results in two equations: \(p+\frac{1}{2}=\frac{3}{2}\) and \(p+\frac{1}{2}=-\frac{3}{2}\).

Step 3: Solve for p in Each Equation

Now solve each equation separately: \(p + \frac{1}{2} = \frac{3}{2}\)1. Subtract \(\frac{1}{2}\) from both sides: \(p = \frac{3}{2} - \frac{1}{2}\).2. Simplify the right side: \(p = 1\). For \(p + \frac{1}{2} = -\frac{3}{2}\)1. Subtract \(\frac{1}{2}\) from both sides: \(p = -\frac{3}{2} - \frac{1}{2}\).2. Simplify the right side: \(p = -2\).

Step 4: Verify the Solutions

Substitute \(p = 1\) and \(p = -2\) back into the original equation to ensure both work.1. For \(p = 1\): \(\big(1+\frac{1}{2}\big)^{2} = \big(\frac{3}{2}\big)^{2} = \frac{9}{4}\).2. For \(p = -2\): \(\big(-2+\frac{1}{2}\big)^{2} = \big(-\frac{3}{2}\big)^{2} = \frac{9}{4}\).Both solutions are valid.

Key Concepts

quadratic formula

Quadratic equations are fundamental in algebra. They take the standard form: \(ax^2 + bx + c = 0\), where \(a, b\), and \(c\) are coefficients.
The quadratic formula is a powerful tool for solving quadratics: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Using the quadratic formula can help find the roots of any quadratic equation. These roots represent the values of \(x\) that make the equation true. Remember, the expression under the square root, \(b^2 - 4ac\), is called the discriminant.

• If it’s positive, there are two real solutions.
• If it’s zero, there’s one real solution.
• If it’s negative, the solutions are complex or imaginary.

In our original problem, however, we used a different method known as the square root principle.

square root principle

The square root principle is a straightforward technique for solving quadratic equations in the form \( (ax + b)^2 = c\).
To utilize the square root principle, isolate the squared term, then take the square root of both sides. Don’t forget that taking the square root introduces two potential solutions: a positive and a negative square root.
In the given problem: \( \left(p + \frac{1}{2}\right)^{2} = \frac{9}{4}\), we take the square root of both sides to get:

• \[p + \frac{1}{2} = \pm \frac{3}{2}\]

This results in two equations: \( p + \frac{1}{2} = \frac{3}{2} \) and \( p + \frac{1}{2} = -\frac{3}{2}\). Solving these will give us the values of \(p\).

simplifying equations

Simplifying equations is an essential algebraic process. It involves reducing equations to their simplest form so they are easier to solve.
In our problem, after applying the square root principle, we had two equations: \( p + \frac{1}{2} = \frac{3}{2} \) and \( p + \frac{1}{2} = -\frac{3}{2}\).
To solve these, follow these steps:

• For \( p + \frac{1}{2} = \frac{3}{2} \), subtract \( \frac{1}{2}\) from both sides to get \( p = 1 \).
• For \( p + \frac{1}{2} = -\frac{3}{2} \), subtract \(\frac{1}{2}\) from both sides to get \( p = -2 \).

These steps illustrate how the equations simplify to provide solutions for \(p\).

verifying solutions

Verification is the final step in solving equations. It ensures that the solutions obtained are correct by substituting them back into the original equation.
For the solutions found in previous steps:

• Substituting \( p = 1 \) into the original equation: \( \left(1 + \frac{1}{2}\right)^{2} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4} \).
• Substituting \( p = -2 \) into the original equation: \( \left(-2 + \frac{1}{2}\right)^{2} = \left(-\frac{3}{2}\right)^{2} = \frac{9}{4} \).

Both solutions satisfy the original equation, confirming they are correct. Verifying results helps avoid errors and ensures confidence in the solutions.