Problem 414 In the following exercises, solv... [FREE SOLUTION] (2024)

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Chapter 9: Problem 414

In the following exercises, solve. $$ 2 \sqrt{8 r+1}-8=2 $$

Short Answer

Expert verified

r = 3

Step by step solution


- Divide by 2

Next, divide both sides of the equation by 2 to isolate the square root even further: \(\frac{2 \sqrt{8r + 1}}{2} = \frac{10}{2}\) Which simplifies to: \(\sqrt{8r + 1} = 5\)


- Square both sides

Square both sides to eliminate the square root: \((\sqrt{8r + 1})^2 = 5^2\) Which results in: \(8r + 1 = 25\)


- Solve for r

Subtract 1 from both sides to isolate the term with r: \(8r + 1 - 1 = 25 - 1\) Which simplifies to: \(8r = 24\) Finally, divide by 8: \(\frac{8r}{8} = \frac{24}{8}\) So: \(r = 3\)

Key Concepts

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

Isolating the Square Root

To solve radical equations, the first important step is isolating the square root. This means we try to get the square root term by itself on one side of the equation.
In our example, the equation is: \[ 2 \sqrt{8r + 1}-8=2 \]We start by adding 8 to both sides to move the negative 8 away from the square root term:
\[ 2 \sqrt{8r + 1} - 8 + 8 = 2 + 8 \]This simplifies to:
\[ 2\sqrt{8r + 1} = 10 \]Now, we can easily work with the square root term. We divide by 2 to further isolate the square root:
\[ \frac{2\sqrt{ 8r + 1 }}{2} = \frac{10}{2} \]Which simplifies to:
\[ \sqrt{ 8r + 1 } = 5 \]
By isolating the square root, we make the equation more manageable for the next steps.

  • Always move terms without the square root to one side first
  • To simplify, divide or multiply if needed
  • Ensure the square root stands alone
Squaring Both Sides

After isolating the square root, the next key step is squaring both sides of the equation. This removes the square root, making it easier to solve for the variable.
In our example, we have:
\[ \sqrt{8r + 1} = 5 \]Squaring both sides will eliminate the square root:
\[ (\sqrt{8r + 1})^2 = 5^2 \]This simplifies to:
\[ 8r + 1 = 25 \]When squaring both sides, always ensure your original square root is isolated. Be careful, as squaring can introduce extraneous solutions. Always verify your solutions in the original equation.

  • Isolate the square root before squaring
  • Square both sides accurately
  • Always check for extraneous solutions
Solving for a Variable

Now that we've squared both sides and no longer have the square root, our next goal is solving for the variable. Let's look at our current equation:
\[ 8r + 1 = 25 \]First, we isolate the term with the variable by subtracting 1 from both sides:
\[ 8r + 1 - 1 = 25 - 1 \]Which simplifies to:
\[ 8r = 24 \]To find the value of r, we then divide both sides by 8:
\[ \frac{8r}{8} = \frac{24}{8} \]
This final step gives us:
\[ r = 3 \]Solving for a variable often involves simple arithmetic after isolating terms. Always perform inverse operations systematically and revisit earlier steps if your solution doesn’t seem correct.

  • Isolate the variable term through addition/subtraction
  • Divide or multiply as necessary
  • Check your solution in the original equation

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Problem 414 In the following exercises, solv... [FREE SOLUTION] (3)

Most popular questions from this chapter

In the following exercises, solve. $$ \sqrt{5 x-6}=8 $$In the following exercises, simplify and rationalize the denominator. $$ \sqrt{\frac{8}{45}} $$In the following exercises, simplify by rationalizing the denominator. $$ \frac{\sqrt{5}}{\sqrt{y}-\sqrt{7}} $$In the following exercises, simplify by rationalizing the denominator. (a) \(\frac{6}{6+\sqrt{5}}\) (b) \(\frac{5}{4-\sqrt{11}}\)In the following exercises, simplify and rationalize the denominator. $$ \frac{10}{3 \sqrt{10}} $$
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Problem 414 In the following exercises, solv... [FREE SOLUTION] (2024)
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