Problem 31 - Elementary Algebra

Question

Solve using the Square Root Property. \(2 r^{2}=32\)

Step-by-Step Solution

Answer

r = 4, -4

Step 1: Isolate the Quadratic Term

First, isolate the term with the variable squared, which in this case is the term with \(r^2\). To do this, divide both sides of the equation by 2: \[ \frac{2r^2}{2} = \frac{32}{2} \] Simplifying, we get: \[ r^2 = 16 \]

Step 2: Apply the Square Root Property

Next, apply the square root property which states that if \ r^2 = k \ then \ r = \pm \sqrt{k} \. In our case, we have: \[ r = \pm \sqrt{16} \]

Step 3: Simplify the Square Root

Finally, simplify the square root: \[ r = \pm 4 \] This means there are two solutions: \[ r = 4 \ and \ r = -4 \]

Key Concepts

Square Root Property

The Square Root Property is a useful tool for solving quadratic equations. It states that if you have an equation where a variable squared equals a constant, like \( r^2 = k \), you can find the solutions for the variable by taking the square root of both sides.
This leads to two possible values: one positive and one negative, as any number squared will give the same positive value.
For example, if \( r^2 = 16 \), taking the square root of both sides gives: \[ r = \pm \sqrt{16} \] which simplifies to:

• \( r = 4 \)
• \( r = -4 \)

Using this property helps to quickly find both solutions to the equation.

Isolating the variable

To solve the quadratic equation efficiently, we first need to isolate the term containing the variable.
This means getting \( r^2 \) by itself on one side of the equation.
In the exercise, we start with \( 2r^2 = 32 \).
We need to isolate \( r^2 \). To do this:

• Divide both sides by 2: \[ \frac{2r^2}{2} = \frac{32}{2} \]
• Simplify to get: \[ r^2 = 16 \]

This step ensures that the equation is in a form where we can easily apply the Square Root Property.

Simplifying radicals

After applying the Square Root Property, we often end up with a radical, which needs to be simplified.
Simplifying radicals makes it easier to understand and work with the solutions.
In the exercise, \( \sqrt{16} \) needs to be simplified.
Since 16 is a perfect square (4 x 4), we can write: \[ \sqrt{16} = 4 \]
Hence, taking the square root of 16 gives us two solutions:

• \( 4 \)
• \( -4 \)

Remember, a radical is simplified when it's written in its simplest form, with any possible perfect squares taken out of the radical symbol.
Always look for perfect squares (like 4, 9, 16) within the number under the radical to simplify it correctly.