Problem 53 - Algebra for College Students

Question

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$4 a(a+1)=3$$

Step-by-Step Solution

Answer

The solutions are \(a = \frac{1}{2}\) and \(a = -\frac{3}{2}\).

Step 1: Expand the Equation

First, expand the equation by distributing the \(4a\) across the \((a+1)\) term. This gives:\[4a^2 + 4a = 3\]

Step 2: Set the Equation to Zero

Subtract \(3\) from both sides of the equation to set the equation to zero:\[4a^2 + 4a - 3 = 0\]

Step 3: Factor the Quadratic Equation

Now, factor the quadratic equation. We look for two numbers that multiply to \(-12\) (coefficient of \(4a^2\) multiplied by \(-3\)) and add to \(4\) (coefficient of \(a\)). These numbers are \(6\) and \(-2\). Rewrite the middle term using these numbers:\[4a^2 + 6a - 2a - 3 = 0\]Now, factor by grouping:\[2a(2a + 3) - 1(2a + 3) = 0\]Factor out \((2a + 3)\):\[(2a - 1)(2a + 3) = 0\]

Step 4: Solve the Factored Equation

Set each factor equal to zero and solve for \(a\):1. \(2a - 1 = 0\) \[ 2a = 1 \ a = \frac{1}{2} \]2. \(2a + 3 = 0\) \[ 2a = -3 \ a = -\frac{3}{2} \]

Step 5: Verify the Solutions

Verify by substituting both solutions back into the original equation:1. \(a = \frac{1}{2}\) \[ 4\left(\frac{1}{2}\right)\left(\frac{1}{2} + 1\right) = 3 \ 4\left(\frac{1}{2}\right)\left(\frac{3}{2}\right) = 3 \ 3 = 3 \] (True)2. \(a = -\frac{3}{2}\) \[ 4\left(-\frac{3}{2}\right)\left(-\frac{3}{2} + 1\right) = 3 \ 4\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right) = 3 \ 3 = 3 \] (True)Both solutions satisfy the original equation.

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra. They are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not equal to zero. The highest power of the variable is 2, which is why they are referred to as 'quadratic.'
These equations can model various real-life phenomena, ranging from projectile motion to economics. Understanding how to solve them is crucial for progressing in algebra and other mathematical fields.
There are different methods to solve quadratics, including factoring, completing the square, and using the quadratic formula. In this scenario, we'll focus on solving by factoring, a technique that involves expressing the quadratic in a product form.

Factoring by Grouping

Factoring by grouping is a method used to factor quadratic equations when simple factoring is not immediately obvious.
This technique involves rearranging terms and grouping them into smaller expressions that can be factored more easily.
For example, consider the quadratic \(4a^2 + 4a - 3 = 0\). The challenge here is to identify two numbers that multiply to \(-12\) (the product of \(4\) and \(-3\)) and add to \(4\) (the coefficient of the \(a\) term). These numbers are \(6\) and \(-2\).
You then rewrite the quadratic as \(4a^2 + 6a - 2a - 3 = 0\). Group the terms to factor out common factors: \(2a(2a + 3) - 1(2a + 3) = 0\).
Now, factor out \((2a + 3)\) to get \((2a - 1)(2a + 3) = 0\). This is how factoring by grouping simplifies the equation and uncovers the solutions.

Algebraic Solutions

Once a quadratic is factored, finding the solutions becomes straightforward. It involves applying the Zero Product Property, which states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
For the factors \((2a - 1)(2a + 3) = 0\), we find the solutions by setting each factor to zero:

• For \(2a - 1 = 0\), we solve \(2a = 1\), giving us \(a = \frac{1}{2}\).
• For \(2a + 3 = 0\), we solve \(2a = -3\), resulting in \(a = -\frac{3}{2}\).

These solutions satisfy the original quadratic equation when substituted back, demonstrating their accuracy.

Verification of Solutions

Verifying solutions is a critical step to ensure their correctness. This is achieved by substituting the found values back into the original equation and checking if the equation holds true.
For \(a = \frac{1}{2}\), substituting into \(4a(a+1)=3\) gives \(4\left(\frac{1}{2}\right)\left(\frac{1}{2} + 1\right) = 3\), which simplifies to \(3 = 3\). This proves it is a valid solution.
Likewise, for \(a = -\frac{3}{2}\), substituting back also gives \(3 = 3\), confirming this as another valid solution.
By verifying, we confirm the solutions accurately satisfy the quadratic equation, ensuring no calculation errors have occurred during solving.