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Chapter 11: Problem 16

Solve. $$ c^{2}-8=0 $$

Short Answer

Expert verified
c = \pm 2\sqrt{2}

Step by step solution

01

Set the equation to isolate the variable term

Start by adding 8 to both sides of the equation: \( c^{2} - 8 + 8 = 0 + 8 \)
02

Simplify the equation

This simplifies to: \( c^{2} = 8 \)
03

Solve for the variable

Take the square root of both sides of the equation to solve for \( c \): \( c = \pm \sqrt{8} \). Simplify the square root: \( c = \pm 2\sqrt{2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isolating the Variable
To solve quadratic equations, it's crucial to isolate the variable. In our exercise, the goal is to get the term with the variable, in this case, \(c^2\), by itself on one side of the equation. We start with:
\(c^2 - 8 = 0\).
To isolate \(c^2\), we add 8 to both sides:
\(c^2 - 8 + 8 = 0 + 8\).
It's important to keep the equation balanced, so whatever operation you do to one side, you must do to the other.
After adding 8, the equation becomes:
\(c^2 = 8\).
Now, \(c^2\) is isolated, ready for further steps.
Simplifying Equations
Simplifying equations ensures that they are easier to solve. After isolating the variable term in our exercise, we simplified:
\(c^2 - 8 + 8 = 8\).
Simplify the left side:
\(c^2\) remains, as \(-8 + 8\) cancels out.
The equation simplifies to:
\(c^2 = 8\).
This step is crucial because it takes a complex expression and simplifies it into a more manageable form. Once simplified, the problem is closer to a solution and less intimidating.
Square Roots
When a variable is squared, like \(c^2\) in our equation, finding the square root of both sides helps solve for the variable. For our equation:
\(c^2 = 8\).
We take the square root of both sides to solve for \(c\):
\(c = \pm \sqrt{8}\). The square root symbol (鈭�) indicates the principal (positive) and its opposite (negative) root. Thus, \(c\) can be both positive and negative.
Let's simplify \(\sqrt{8}\). We know \(8 = 4 \cdot 2\) and that \(\sqrt{4} = 2\). So, \(c = \pm 2\sqrt{2}\).This gives us two solutions:
\(c = 2\sqrt{2}\) and
\(c = -2\sqrt{2}\).
By understanding square roots, you can solve quadratic equations and find all possible solutions for the variable.

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