Problem 62
Question
Find all real number solutions for each equation. $$3 x^{2}-108=0$$
Step-by-Step Solution
Verified Answer
The solutions are \(x = 6\) and \(x = -6\).
1Step 1: Isolate the Quadratic Term
Start with the equation: \(3x^2 - 108 = 0\). Our first goal is to isolate the term \(x^2\). To do this, add 108 to both sides of the equation: \(3x^2 = 108\).
2Step 2: Solve for \(x^2\)
Now we have the equation \(3x^2 = 108\). To solve for \(x^2\), divide both sides by 3: \(x^2 = \frac{108}{3}\). Simplifying the right side gives: \(x^2 = 36\).
3Step 3: Take the Square Root
To find \(x\), we take the square root of both sides of the equation \(x^2 = 36\). Remember, the square root of a number can be positive or negative. Thus, \(x = \sqrt{36}\) or \(x = -\sqrt{36}\).
4Step 4: Simplify the Square Root
Calculate \(\sqrt{36}\) which is 6. So, \(x = 6\) or \(x = -6\).
5Step 5: Conclusion
The real number solutions to the equation are \(x = 6\) and \(x = -6\).
Key Concepts
Real Number SolutionsQuadratic Term IsolationSquare Root MethodEquation Solving Steps
Real Number Solutions
In the realm of mathematics, when we talk about real number solutions, we are referring to answers to equations that are real numbers. Real numbers are those found on the number line, which include all the integers, fractions, finite decimals, and irrational numbers. Importantly, they do not include complex numbers involving imaginary units. For quadratic equations like the one you're dealing with, the solutions can either be real or complex. In our example problem, since the solutions are both integers, they fall under the category of real solutions. This means they are values that can be plotted on the number line and concretely represent the solutions of our quadratic equation.
Quadratic Term Isolation
Quadratic term isolation is a fundamental step when solving equations like the given one. The initial equation is \(3x^2 - 108 = 0\), and our goal is to isolate the squared term \(x^2\) on one side. This process involves performing algebraic operations to get \(x^2\) on its own.
- The first operation we use is adding 108 to both sides. This step cancels out the -108, resulting in \(3x^2 = 108\).
- Afterwards, to further isolate \(x^2\), we divide the entire equation by 3. This step removes the coefficient of \(x^2\), simplifying the equation to \(x^2 = 36\).
Square Root Method
The square root method is an efficient technique for solving equations where the quadratic term has been isolated. Once you have the equation in the form \(x^2 = a\), you can apply the square root method. This involves taking the square root of both sides to solve for \(x\).
- Remember that the square root of any positive number has two values: positive and negative. This is because both a positive and a negative number squared result in the same positive value. For example, \(x^2 = 36\) leads to \(x = \sqrt{36}\) or \(x = -\sqrt{36}\).
- By calculating \(\sqrt{36}\), we find that \(x\) equals 6 or -6. These operations are straightforward but crucial as they provide the solutions to the equations.
Equation Solving Steps
Solving quadratic equations can be systematic when broken down into clear steps. Let’s recap the steps involved using our given equation as an exemplar:
- Step 1: Start with the Original Equation. Our task begins with \(3x^2 - 108 = 0\).
- Step 2: Quadratic Term Isolation. Adjust the equation to get \(x^2\) by itself: \(3x^2 = 108\), then \(x^2 = 36\).
- Step 3: Apply the Square Root Method. Take the square root of both sides: \(x = \sqrt{36}\) and \(x = -\sqrt{36}\).
- Step 4: Simplify and Find the Solutions. Simplifying the square roots, we get \(x = 6\) and \(x = -6\).
- Conclusion: Confirm that the solutions satisfy the original equation, verifying our steps.
Other exercises in this chapter
Problem 62
Set up an equation and solve each problem. The sum of the areas of two circles is \(65 \pi\) square feet. The length of a radius of the larger circle is 1 foot
View solution Problem 62
Should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using
View solution Problem 62
Factor by grouping. $$x^{2}-2 x+5 x-10$$
View solution Problem 62
Find each indicated product. Remember the shortcut for multiplying binomials and the other special patterns we discussed in this section. $$\left(x^{2}-x+6\righ
View solution