Problem 77 - College Algebra

Question

Solve each equation in Exercises \(73-98\) by the method of your choice. \(3 x^{2}=60\)

Step-by-Step Solution

Answer

The solutions to the equation \(3 x^{2}=60\) are \(x = 2\sqrt{5}\) and \(x = -2\sqrt{5}\)

Step 1: Isolate the square term

The first step is to isolate \(x^2\) by dividing through by the coefficient of the x term, which is 3. This gives us the equation \[x^{2} = 60/3 = 20\]

Step 2: Apply the square root principle

To isolate x, we apply square root to both sides of the equation. This gives two solutions, because the square root of a number can be both positive and negative. So, \[x = \sqrt{20}\] and \[x = -\sqrt{20}\]

Step 3: Simplify the square root

Square root of 20 can be simplified by factoring 20 as \(4*5\) and taking square root of 4 which gives 2 as a result. So, \[x = 2\sqrt{5}\] and \[x = -2\sqrt{5}\]

Key Concepts

Solving Quadratic Equations

To solve a quadratic equation, like the one given in the original problem, it's important to identify the standard form first, which is \(ax^2 + bx + c = 0\). In the example \(3x^2 = 60\), the equation can be rearranged and simplified to isolate terms and make the process easier. Here, there is no \(bx\) term, which simplifies our approach even further.
When dealing with equations that are straightforward like this one, we focus on isolating \(x^2\) so it's easier to solve by applying mathematical operations.
In this problem, the key is to first divide every term by 3, the coefficient of \(x^2\). This essential step makes the problem manageable, turning it into \(x^2 = 20\). With \(x^2\) isolated, we are set to use the square root principle, which is a straightforward method to handle such equations when the \(bx\) term is absent or when dealing with pure quadratic expressions.

Square Root Principle

The square root principle is a powerful tool in algebra for solving quadratic equations, especially those lacking an \(x\) term (or when it's set to zero). This principle tells us that if \(x^2 = a\), then \(x\) could be both \(\sqrt{a}\) and \(-\sqrt{a}\).
In the example \(x^2 = 20\), applying this principle involves taking the square root of both sides, introducing a key concept:

• Introduce both the positive and negative roots, hence, \(x = \sqrt{20}\) and \(x = -\sqrt{20}\).
• This shows the symmetry in quadratic equations, as solutions exist on both sides of the number line.

Understanding this principle allows us to effectively find solutions to square-based problems, setting the foundation to further simplify solutions when necessary.

Simplifying Square Roots

Simplifying square roots can often transform a complex-looking solution into a much cleaner one. This includes recognizing and extracting perfect squares from the number under the square root, known as the radicand.
In our example of \(\sqrt{20}\), we apply simplification by breaking down the radicand using factorization:

• First, identify the largest perfect square within 20, specifically 4, which is beneficial since \(4 \times 5 = 20\).
• Next, consider the square root of this perfect square, \(\sqrt{4} = 2\).
• Thus, \(\sqrt{20}\) simplifies to \(2\sqrt{5}\), allowing us to express the roots as \(x = 2\sqrt{5}\) and \(x = -2\sqrt{5}\).

Through simplification, you not only clarify the solution but also make it easier for further mathematical operations, should they be needed. Recognizing and handling square roots efficiently aids in handling various quadratic equations in algebra.