/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Solve each equation. If it has n... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each equation. If it has no solution, write "no solution." $$ 5 \sqrt{x}=25 $$

Short Answer

Expert verified
The solution is \( x = 25 \).

Step by step solution

01

Isolate the Square Root

First, isolate the square root on one side of the equation. In this case, it is already isolated: \( 5 \sqrt{x} = 25 \)
02

Divide by the Coefficient

Divide both sides of the equation by the coefficient of the square root, which is 5, to get: \( \sqrt{x} = \frac{25}{5} \) \ Simplify the right-hand side: \( \sqrt{x} = 5 \)
03

Square Both Sides

Eliminate the square root by squaring both sides of the equation: \( (\sqrt{x})^2 = 5^2 \) \ This simplifies to: \( x = 25 \)
04

Verify the Solution

Substitute \( x = 25 \) back into the original equation to check if it works: \( 5 \sqrt{25} = 25 \) \ Since \( \sqrt{25} = 5 \), the equation becomes: \( 5 \times 5 = 25 \) \ This is true, so \( x = 25 \) is a valid solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Isolating Square Roots
When solving equations involving square roots, the first thing you need to do is isolate the square root term. Isolating means getting the square root by itself on one side of the equation.
In the problem given, the equation is already isolated: \( 5 \sqrt{x} = 25 \). This means the square root term \( \sqrt{x} \) is by itself on one side, which makes the subsequent steps easier to handle.
Dividing Coefficients
Once the square root term is isolated, the next step is to get rid of any coefficients attached to it. Coefficients are numbers that multiply the variable or square root.
In our example, the term \( 5 \sqrt{x} \) has a coefficient of 5.
To eliminate this 5 and simplify the equation, you divide both sides by 5:
\[ \sqrt{x} = \frac{25}{5} \]
After simplifying, you get:
\[ \sqrt{x} = 5 \]
Squaring Both Sides
Next, to get rid of the square root, you square both sides of the equation. Squaring means to raise a number to the power of 2.
For our example, \(\root{x} = 5 \):
\[ (\root{x})^2 = 5^2 \]
This step is key because squaring the square root will cancel out the square root, leaving you with the variable alone.
Simplifying further, you get: \[ x = 25 \]
Verifying Solutions
The final step is to verify if the solution you obtained is correct. This is done by substituting the value back into the original equation.
For our example, substitute \x=25 \ into \( 5 \sqrt{x} = 25 \):
\[ 5 \root{25} = 25 \]
Simplifying \sqrt{25} \ gives you 5:
\[ 5 \times 5 = 25 \]
This confirms that our solution \ x = 25 \ is indeed correct. Verifying helps to make sure no mistakes were made during the process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Identify all the pairs of equivalent fractions in this list. \(\frac{1}{3} \quad \frac{7}{12} \quad \frac{2}{9} \quad \frac{8}{10} \quad \frac{28}{48} \quad \frac{20}{25} \quad \frac{6}{27} \quad \frac{12}{36}\)

A particular tennis tournament begins with 64 players. If a player loses a single match, he or she is knocked out of the tournament. After one round, only 32 players remain; after two rounds, only 16 remain; and so on. Six students have conjectured a formula to describe the number of players remaining, \(p,\) after \(r\) rounds. Which rule or rules are correct? For each rule you think is correct, show how you know. \(\bullet\) Terrill: \(p=\frac{64}{2^{r}}\) \(\bullet\) Mi-Yung: \(p=64 \cdot 2^{-r}\) \(\bullet\) Antonia: \(p=64 \cdot \frac{1}{2^{r}}\) \(\bullet\) Peter: \(p=64 \cdot\left(\frac{1}{2}\right)^{r}\) \(\bullet\) Damon: \(p=64 \cdot 0.5^{r}\) \(\bullet\) Tamika: \(p=64 \cdot(-2)^{r}\)

Rewrite each expression as simply as you can. $$\left(4 x^{-2}\right)^{6}$$

For the linear equations in Exercises \(29-31,\) answer Parts a and b. a. What is the constant difference between the \(y\) values as the \(x\) values increase by 1\(?\) b. What is the constant difference between the \(y\) values as the \(x\) values decrease by 3\(?\) $$ y=\frac{x}{2} $$

Julia found three Web sites that sell 4-inch-square stickers of her favorite band’s logo. The three sites sell the stickers for different prices, and charge different amounts for shipping. Site \(1 :\) Stickers are 75 each; shipping is \(\$ 4\) for any size order. Site \(2 :\) Stickers are 60\(\phi\) each; shipping is \(\$ 5.50\) for any size order. Site \(3 :\) Stickers are \(\$ 1.25 ;\) shipping is included. a. For each site, write an equation to represent the charge \(C\) for ordering any number of stickers \(s\) b. Graph your three equations on axes like these. Label each graph with its site number. c. Use your graph to answer this question: If Julia wants to order 16 stickers, which site will charge her the least? d. Use your graph to answer this question: If Julia wants to order 10 stickers, which site will charge her the least?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.