/*! 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 55 Solve each equation. $$ \sqr... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 55

Solve each equation. $$ \sqrt{x+3}-1=4 $$

Short Answer

Expert verified
The solution for x in the equation \(\sqrt{x+3} - 1 = 4\) is \(x = 22\).

Step by step solution

01

Isolate the square root

First, add 1 to both sides of the equation to isolate the square root on one side. This leaves us with \(\sqrt{x+3} = 5\)
02

Remove the square root

To remove the square root, square both sides of the equation. Squaring \(\sqrt{x+3}\) leaves us with \(x + 3\). Squaring 5 gives 25. Therefore, we now have the equation \(x + 3 = 25\)
03

Solve for x

Finally, subtract 3 from both sides in order to isolate the variable \(x\). This gives us the solution \(x = 22\). Remember to check your solution by substituting it back into the original equation to ensure it holds true.

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Key Concepts

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

Square Root Equations
Square root equations are those that include a square root expression, such as \( \sqrt{x+3}-1=4 \). To solve these equations, it's crucial first to remove or isolate the square root term. This involves algebraic manipulation and sometimes includes squaring both sides of the equation, which can affect the solution.
Remember, each step should maintain the equation's balance to ensure accurate results. Working carefully is key to avoid errors.
Isolating Variables
The process of isolating variables involves strategically rearranging an equation to get the variable by itself on one side.
  • **Step 1:** Look at what operations are affecting the variable.
  • **Step 2:** Perform opposite or inverse operations to eliminate these effects.
For our example, \(\sqrt{x+3} - 1 = 4\), you need to add 1 to both sides to isolate the square root term. This essential step simplifies the equation to \(\sqrt{x+3} = 5\), making it easier to find \(x\) in further steps.
Checking Solutions
Checking solutions is a vital part of solving equations. It ensures your answer is correct and can prevent logical errors.
Once you have found a potential solution, substitute it back into the original equation. For example, substitute \(x = 22\) back into \(\sqrt{x+3} - 1 = 4\):
  • Plug in 22 for \(x\): \(\sqrt{22+3} - 1 = 4\).
  • Calculate the square root: \(\sqrt{25} - 1 = 4\).
  • Simplify: 5 - 1 = 4, which holds true.
This means our solution is verified as correct.
Algebraic Manipulation
Algebraic manipulation involves using basic algebraic principles to rearrange and solve equations. This technique often encompasses several sub-steps.
  • Scheduling logical operations to simplify the equation.
  • Utilizing inverse operations, like addition to cancel subtraction.
  • Squaring both sides to remove a square root.
In the given problem, start by isolating the square root. Then, square the isolated equation to transition from \(\sqrt{x+3} = 5\) to \(x+3 = 25\).
Finally, further isolate \(x\) by subtracting 3, leading to \(x = 22\). These methods are fundamental in resolving complex algebraic equations.

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Most popular questions from this chapter

What is the product of \(\frac{4 x^{2}-1}{2 x^{2}-5 x-3}\) and \(\frac{x^{2}-6 x+9}{2 x^{2}+5 x-3} ?\) $$ \begin{array}{ll}{\text { A. } 1} & {\text { B. } x-3} \\ {\text { C. } x+3} & {\text { D. } \frac{x-3}{x+3}}\end{array} $$

Solve each equation. Check your answer. $$ \frac{2}{2 x-1}=\frac{x}{3} $$

Use this information for Exercises \(53-58\) . Bag 1 contains 5 red marbles, 1 blue marble, 3 yellow marbles, and 2 green marbles. Bag 2 contains 1 red pencil, 3 red pens, 2 blue pencils, and 5 blue pens. One item is drawn from bag 2 . What is the probability that it is a pen or a red pencil?

A standard number cube is tossed. Find each probability. \(P(\text { odd or greater than } 2)\)

Use the fact that \(\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b} \div \frac{c}{d}\) to simplify each rational expression. State any restrictions on the variables. $$ \frac{\frac{8 x^{2} y}{x+1}}{\frac{6 x y^{2}}{x+1}} $$
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