Problem 82 - Algebra for College Students

Question

Find all real or imaginary solutions to each equation. Use the method of your choice. $$5 w^{2}-3=0$$

Step-by-Step Solution

Answer

The solutions are \( w = \pm \frac{\sqrt{15}}{5} \).

Step 1: Isolate the quadratic term

Add 3 to both sides of the equation to isolate the term containing the variable. \[ 5w^2 - 3 + 3 = 0 + 3 \] This simplifies to: \[ 5w^2 = 3 \]

Step 2: Solve for the squared term

Divide both sides by 5 to solve for \( w^2 \). \[ \frac{5w^2}{5} = \frac{3}{5} \] This simplifies to: \[ w^2 = \frac{3}{5} \]

Step 3: Take the square root of both sides

Take the square root of both sides to solve for \( w \). Remember to consider both the positive and negative square roots. \[ w = \pm \sqrt{\frac{3}{5}} \] Since \( \sqrt{\frac{3}{5}} = \frac{\sqrt{3}}{\sqrt{5}} \), we rationalize the denominator: \[ w = \pm \frac{\sqrt{3}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \pm \frac{\sqrt{15}}{5} \]

Key Concepts

Isolating the Variable

Let's start with the first step: isolating the variable. In many equations, especially quadratic ones, isolating the variable is crucial to simplifying the problem.
For the equation given, we need to get the term containing the variable on one side of the equation. Here's how we do it:

• We begin with the equation: \(5w^2 - 3 = 0\)
• To isolate the quadratic term, we add 3 to both sides:

\[5w^2 - 3 + 3 = 0 + 3\]

By adding 3, we eliminate the -3 on the left side:
\[ 5w^2 = 3 \]
Isolating the variable helps keep our equation neat and sets us up for the next steps. If the term containing the variable is not alone, we wouldn't be able to simplify or solve the equation properly.

Taking the Square Root

Now that we have isolated the variable, let's deal with the squared term. We need to simplify further to solve for \(w\).
Our current equation looks like this: \( 5w^2 = 3 \)
To remove the coefficient 5 from the variable term, we divide both sides by 5:
\[ \frac{5w^2}{5} = \frac{3}{5} \]
This simplifies to: \[ w^2 = \frac{3}{5} \]
Next, we take the square root of both sides to solve for \(w\). Remember that whenever you take the square root of a number, you must consider both the positive and negative roots. This gives us:
\[ w = \pm \sqrt{\frac{3}{5}} \]
Taking the square root is a critical step in solving quadratic equations as it allows us to move from \( w^2 \) to \( w \), leading us closer to the final solution.

Rationalizing Denominators

Finally, we need to deal with the square root we obtained. Currently, we have:
\[ w = \pm \sqrt{\frac{3}{5}} \]
Let's simplify this by rationalizing the denominator. Rationalizing the denominator means removing any square roots from the bottom of a fraction.
We start by expressing \( \sqrt{\frac{3}{5}} \) as \(\frac{\sqrt{3}}{\sqrt{5}} \). To rationalize the denominator, multiply both the numerator and the denominator by \( \sqrt{5} \)
\[ w = \pm\frac{\sqrt{3}}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}} \]
After multiplying, we get:
\[ w = \pm\frac{\sqrt{15}}{5} \]
Rewriting the expression in this way gives us the final solutions for the equation. By rationalizing the denominator, we ensure our answer is in the simplest form.