Problem 87
Question
Solve equation by the method of your choice. $$ 3 x^{2}=60 $$
Step-by-Step Solution
Verified Answer
The solutions to the equation are \(x=2*\sqrt{5}\) and \(x=-2*\sqrt{5}\)
1Step 1: Divide both sides by 3
First divide both sides of the equation by 3, to isolate \(x^{2}\) on one side. This gives us \[x^{2} = \frac{60}{3} = 20\]
2Step 2: Square root both sides
Next take square root of both sides to solve for x. Remember there are two possible solutions - a positive and a negative square root. So, the solution would be \[x=\sqrt{20}, -\sqrt{20}\]
3Step 3: Simplify the square root
The square root of 20 can be simplified as it's equivalent to \(2*\sqrt{5}\), so simplify your both values of x as x= \(2*\sqrt{5}, -2*\sqrt{5}\)
Key Concepts
Square RootSimplifying Square RootsPositive and Negative Solutions
Square Root
The square root is a fundamental concept in mathematics. It refers to a number that, when multiplied by itself, results in the original number. The symbol for square root is \( \sqrt{} \). So, if you have \( x = \sqrt{20} \), it means you're looking for a number that multiplied by itself gives 20.
When taking the square root of a number, remember that there can be two solutions: one that is positive and another that is negative. This is because both \( 4 \times 4 = 16 \) and \( -4 \times -4 = 16 \).
In our exercise, after dividing both sides by 3, we are left with \( x^2 = 20 \). To solve for \( x \), we take the square root of both sides, giving us \( x = \pm \sqrt{20} \) which is the point where we introduce the next concept—positive and negative solutions.
When taking the square root of a number, remember that there can be two solutions: one that is positive and another that is negative. This is because both \( 4 \times 4 = 16 \) and \( -4 \times -4 = 16 \).
In our exercise, after dividing both sides by 3, we are left with \( x^2 = 20 \). To solve for \( x \), we take the square root of both sides, giving us \( x = \pm \sqrt{20} \) which is the point where we introduce the next concept—positive and negative solutions.
Simplifying Square Roots
Simplifying square roots involves expressing the square root in its simplest form. This can make calculations easier and results more understandable.
To simplify a square root, you need to find its prime factors. For \( \sqrt{20} \), break down 20 into its prime factors: \( 20 = 2^2 \times 5 \). Then you use these factors to simplify the square root:
This simplification is a critical step because it allows for more precise and easily readable solutions in your calculations.
To simplify a square root, you need to find its prime factors. For \( \sqrt{20} \), break down 20 into its prime factors: \( 20 = 2^2 \times 5 \). Then you use these factors to simplify the square root:
- Identify pairs of prime factors: \( 2 \times 2 \) are a pair.
- For every pair, one number comes out of the square root. So, \( \sqrt{20} = \sqrt{2^2 \times 5} = 2\sqrt{5} \).
This simplification is a critical step because it allows for more precise and easily readable solutions in your calculations.
Positive and Negative Solutions
When solving an equation with a square root, such as \( x^2 = 20 \), it’s important to account for both positive and negative solutions. This is because any positive number squared is equal to the magnitude of its negative counterpart squared.
For example, if \( x = \sqrt{20} \), then \(-x = -\sqrt{20} \) is also a valid solution for \( x^2 = 20 \). This dual solution arises from the fact that squaring either the positive \( \sqrt{20} \) or the negative \( -\sqrt{20} \) results in the same value.
In our simplified solution \( x = 2\sqrt{5} \), we similarly have two solutions:
For example, if \( x = \sqrt{20} \), then \(-x = -\sqrt{20} \) is also a valid solution for \( x^2 = 20 \). This dual solution arises from the fact that squaring either the positive \( \sqrt{20} \) or the negative \( -\sqrt{20} \) results in the same value.
In our simplified solution \( x = 2\sqrt{5} \), we similarly have two solutions:
- \( x = 2\sqrt{5} \)
- \( x = -2\sqrt{5} \)
Other exercises in this chapter
Problem 86
Evaluate \(x^{2}-x\) for the value of \(x\) satisfying $$2(x-6)=3 x+2(2 x-1)$$
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In Exercises 59–94, solve each absolute value inequality. $$ 5>|4-x| $$
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Will help you prepare for the material covered in the next section. If 6 is substituted for \(x\) in the equation $$ 2(x-3)-17=13-3(x+2) $$ is the resulting sta
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Find all values of \(x\) satisfying the given conditions.. $$y=x+\sqrt{x+5} \text { and } y=7$$
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