Question

\(\frac{1}{4} y^2=x\)

Short answer

The solutions are \(y=\sqrt{4x}\) or \(y=-\sqrt{4x}\)

Step-by-step solution

Rearranging the equation

Rearrange the given equation for y. It would become: \(y^2=4x\)

Applying the square root

Applying the square root to both sides to isolate y, it would become: \(y=\sqrt{4x}\) or \(y=-\sqrt{4x}\) as squaring in the equation might have introduced an extraneous solution.

Key concepts

Rearranging Equations

Rearranging equations is a fundamental skill in algebra that helps to simplify and solve for a specific variable. In the equation given, \( \frac{1}{4} y^2 = x \), we need to solve for \( y \).
The goal is to isolate \( y^2 \) on one side. To do this, multiply both sides of the equation by 4. This gets rid of the fraction:

• Initial equation: \( \frac{1}{4} y^2 = x \)
• Multiply both sides by 4: \( y^2 = 4x \)

By following these steps, we've rearranged the equation to make \( y^2 \) the subject. Rearranging is essential because it allows us to work toward finding the value of \( y \) by isolating its term.

Square Roots

After rearranging the equation to \( y^2 = 4x \), our next focus is on using the square root. Taking the square root of both sides of the equation helps solve for \( y \), which means finding the original number when squared.
When you take the square root of \( y^2 \), it gives \( y \), but there is a crucial detail: both positive and negative roots should be considered. This is because both \( y = 2 \) and \( y = -2 \) will result in \( y^2 = 4 \). So the equation becomes:

• \( y = \sqrt{4x} \)
• \( y = -\sqrt{4x} \)

Always remember: whenever you take the square root, check for two possible solutions. This two-solution scenario is a key aspect of working with square roots.

Extraneous Solutions

In the process of solving equations, particularly involving square roots, you might encounter extraneous solutions. These are solutions that emerge during manipulations like squaring both sides but are not valid for the original equation.
In the step where we solve \( y^2 = 4x \), both \( y = \sqrt{4x} \) and \( y = -\sqrt{4x} \) are derived. However, it's important to substitute these back into the original equation to verify which, if any, is relevant.

• An extraneous solution doesn't satisfy the original equation.
• Checking each solution ensures they're valid.

Always pay attention to this step to ensure your solutions are correct and applicable to the initial problem you are solving.