Problem 15 - Beginning and Intermediate Algebra

Question

Solve each equation.. $$\left(y+\frac{3}{2}\right)\left(y-\frac{1}{4}\right)=0$$

Step-by-Step Solution

Answer

The short answer to the given equation \(\left(y+\frac{3}{2}\right)\left(y-\frac{1}{4}\right)=0\) is \(y=-\frac{3}{2}\) and \(y=\frac{1}{4}\).

Step 1: Identify the factors

First, we identify the factors of this quadratic equation. From the given equation, we have: \[ \left(y+\frac{3}{2}\right)\left(y-\frac{1}{4}\right)=0 \] So, the factors are \(y+\frac{3}{2}\) and \(y-\frac{1}{4}\).

Step 2: Find the roots

Now we need to find the values of \(y\) for which any of these factors become zero. For that, we solve each factor as follows: 1. Set the first factor to zero and solve for \(y\): \[ y+\frac{3}{2}=0 \] 2. Set the second factor to zero and solve for \(y\): \[ y-\frac{1}{4}=0 \]

Step 3: Solve for y

Solving the equations from step 2, we'll find the values of \(y\). 1. For the first equation, isolate \(y\): \[ y+\frac{3}{2}=0 \] Subtracting \(\frac{3}{2}\) from both sides, we get: \[ y=-\frac{3}{2} \] 2. For the second equation, isolate \(y\): \[ y-\frac{1}{4}=0 \] Adding \(\frac{1}{4}\) to both sides, we get: \[ y=\frac{1}{4} \]

Step 4: Write the final answer

Now, we have found the two roots of the equation which make the equation equal to zero. These roots are: \[ y=-\frac{3}{2}, \ y=\frac{1}{4} \]

Key Concepts

Factoring in Quadratic Equations

Factoring is an essential method used in solving quadratic equations. It involves breaking down the equation into simpler components, known as factors. In this context, each factor is an expression that when multiplied together, gives you the original quadratic expression.

In our given example, the quadratic expression \[(y+\frac{3}{2})(y-\frac{1}{4})=0\]illustrates how factoring is used.

• Each factor here is a linear expression: \(y+\frac{3}{2}\) and \(y-\frac{1}{4}\).
• The equation relies on the zero product property, which states that if the product of two numbers is zero, then at least one of the numbers must be zero.

Factoring helps in reducing the complexity of the equation and allows for straightforward ways to find solutions by setting each factor to zero. This method is especially useful when dealing with simple quadratic expressions like the one given.

Understanding the Roots of Equations

The roots of a quadratic equation are the values of the variable that make the entire equation equal to zero. These are critical because they represent the solution to the problem. In our example, after factoring the equation, we find the roots by setting each factor equal to zero.

This leads us to two simple linear equations:

• \(y+\frac{3}{2}=0\)
• \(y-\frac{1}{4}=0\)

Solving these tells us that the roots are:

• \(y=-\frac{3}{2}\), achieved when the first factor is zero.
• \(y=\frac{1}{4}\), achieved when the second factor is zero.

Roots can also be called solutions or zeros of the equation. Understanding them provides insight into the behavior of the quadratic expression.

Solving Equations Through Factoring

To solve a quadratic equation through factoring, follow a systematic approach. The goal is to find solutions that satisfy the equation, thus making it zero. The process is as follows:

First, ensure that the equation is neatly factored, which often requires rearranging and simplifying. In our exercise, factoring is already done for us. Then, apply the zero product property, meaning if you have \[(a)(b)=0\]at least one of \(a\) or \(b\) must be zero.

• Set each factor equal to zero.
• Solve these resulting equations individually to find possible values of the variable.

By factored equation \[(y+\frac{3}{2})(y-\frac{1}{4})=0\], solving gives the roots

• \(y=-\frac{3}{2}\)
• \(y=\frac{1}{4}\)

This shows factoring not only simplifies the problem but also offers a direct route to finding solutions for quadratic equations.