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Chapter 10: Problem 36

See Example 5. Let \(f(x)=(x+3)^{2} .\) For what value(s) of \(x\) is \(f(x)=7 ?\)

Short Answer

Expert verified
The solutions are \(x = \sqrt{7} - 3\) and \(x = -\sqrt{7} - 3\).

Step by step solution

01

Set the Function Equal to 7

To find the value of \(x\) where \(f(x) = 7\), we start by setting the function equal to 7: \((x+3)^2 = 7\).
02

Take the Square Root of Both Sides

To remove the square, take the square root of both sides: \(\sqrt{(x+3)^2} = \sqrt{7}\). This simplifies to \(x + 3 = \pm \sqrt{7}\).
03

Solve for x in Two Cases

We have two cases to consider due to the \(\pm\) sign: - Case 1: \(x + 3 = \sqrt{7}\) - Case 2: \(x + 3 = -\sqrt{7}\).
04

Solve Each Equation for x

For Case 1, subtract 3 from both sides: \(x = \sqrt{7} - 3\).For Case 2, subtract 3 from both sides: \(x = -\sqrt{7} - 3\).
05

State the Solutions

The solutions to \(f(x) = 7\) are \(x = \sqrt{7} - 3\) and \(x = -\sqrt{7} - 3\).

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Key Concepts

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

Understanding Function Evaluation
Function evaluation is the process of determining the output of a function based on a given input. To evaluate a function, you simply substitute the value of the variable (in this case, "x") into the function expression. For example, if you are given a function such as \(f(x) = (x+3)^2\) and asked to find \(f(x) = 7\), it means you need to determine which values of \(x\) make the function equal to 7.
  • Step 1: Substitute the value. In our example, we set \((x+3)^2 = 7\).
  • Step 2: Solve for the variable. This involves manipulating the equation to find "x".
This process is essential in algebra as it allows you to find specific solutions and comprehend how changes in "x" affect "f(x)".
Applying the Square Root Property
The square root property is a powerful tool to solve equations that involve squares. When an equation involves a squared term, like \((x+3)^2 = 7\), one effective way to solve for "x" is by utilizing this property.For any equation of the form \(a^2 = b\), you can take the square root of both sides to yield \(a = \pm\sqrt{b}\). This means that "a" could be either the positive or negative square root of "b".In our given function \((x+3)^2 = 7\):
  • Take the square root of each side: \(\sqrt{(x+3)^2} = \pm\sqrt{7}\)
  • This reduces to: \(x+3 = \pm\sqrt{7}\)
Utilizing the square root property allows us to tackle problems involving squares easily, providing two potential solutions, which are essential for thorough problem solving.
Mastering Algebraic Manipulation
Algebraic manipulation involves the rearranging and simplification of equations to isolate the desired variable. It is a critical skill in algebra and helps in solving equations efficiently. Let's consider the steps with our specific equation \(x+3 = \pm\sqrt{7}\):Algebraic manipulation will help you isolate "x":
  • Subtract 3 from both sides of the equation to solve for "x":
    - For \(x+3 = \sqrt{7}\), rearrange to get \(x = \sqrt{7} - 3\).
    - For \(x+3 = -\sqrt{7}\), rearrange to get \(x = -\sqrt{7} - 3\).
These simple steps highlight how algebraic manipulation grants you the tools to explore and solve a wide array of mathematical problems, making the task of finding "x" straightforward and intuitive.

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Most popular questions from this chapter

Solve each equation. Approximate the solutions to the nearest hundredth when appropriate. $$ x^{2}+8 x+6=0 $$

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Explain why completing the square on \(x^{2}+5 x\) is more difficult than completing the square on \(x^{2}+4 x\).

Albert Einstein discovered a connection between energy and mass. This relationship (energy equals mass times the velocity of light squared) is expressed in the equation \(E=m c^{2} .\) Solve for \(c\).

Solve each inequality. Write the solution set in interval notation and graph it. $$ x^{2}+4 x+4>0 $$
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