Problem 13 - Elementary Algebra

Question

\(\frac{1}{3} x^{2}=50\)

Step-by-Step Solution

Answer

x \approx \pm 12.247

Step 1: Isolate the Variable Term

Multiply both sides of the equation by 3 to isolate the variable term. \[ \frac{1}{3} x^2 \times 3 = 50 \times 3 \ x^2 = 150 \]

Step 2: Solve for the Variable

Take the square root of both sides of the equation to solve for the variable x. \[ x = \sqrt{150} \ x = \pm \sqrt{150} \ x \approx \pm 12.247 \]

Key Concepts

Isolating the Variable

To solve a quadratic equation like \(\frac{1}{3}x^2 = 50\), the first step is to isolate the variable term. Here, the variable term is \(\frac{1}{3}x^2\).

Isolating the variable means getting \(x^2\) by itself on one side of the equation. This often involves reversing operations that are being performed on the variable. In our problem, \(x^2\) is being multiplied by \(\frac{1}{3}\).

To isolate \(x^2\), we multiply both sides of the equation by 3. This step effectively cancels out the fraction:
\[\frac{1}{3}x^2 \times 3 = 50 \times 3\]
\[x^2 = 150\]

Now, we have successfully isolated \(x^2\), setting us up for the next step in solving the equation.

Multiplying Both Sides

Multiplying both sides of an equation by the same number is a fundamental technique in algebra. This helps keep the equation balanced while simplifying it.

Let's apply this concept to our example:
\(\frac{1}{3}x^2 = 50\).
To get rid of the fraction, we multiply both sides by 3:
\[\frac{1}{3}x^2 \times 3 = 50 \times 3\]
\[x^2 = 150\]

When you multiply both sides by the same number, you must remember a few key points:

• This operation keeps the equation balanced.
• It helps isolate the variable term for easier solving.
• Always perform exactly the same operation on both sides.

Taking Square Roots

Once \((x^2)\) is isolated, solving for \((x)\) requires taking the square root on both sides. Square rooting essentially reverses squaring, but remember: every positive number has two square roots—one positive and one negative.

From our isolated equation \(x^2 = 150\), we get:
\[x = \pm \sqrt{150}\]

This leaves us with two solutions: \[x = \sqrt{150} \text{and}\ x = \- \sqrt{150}\]

Often, we approximate these using decimal values. Using a calculator, we find:
\[x \approx +12.247 \text{and}\ x \approx -12.247\]

The square root step is crucial because it turns the quadratic equation into solvable linear forms. This final step completes our solution process.