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Chapter 10: Problem 32

Solve each radical equation. $$(x-3)^{\frac{1}{2}}+8=6$$

Short Answer

Expert verified
The given equation has no real solutions.

Step by step solution

01

Isolate the Radical

Need to isolate the radical term in the equation. To achieve this, subtract 8 from both sides of the equation. So the equation becomes: \((x-3)^{\frac{1}{2}} = 6 - 8 \) or \((x-3)^{\frac{1}{2}} = -2\)
02

Handle Radical

After isolating the radical, we notice that it is equal to a negative number \( -2 \). It's important to note that the square root of a real number can't be negative. Hence, the equation has no real solutions.
03

Solve for \( x \)

In this step, there's nothing to do, because as we have determined in the last step, there's no real solution for this equation.

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Key Concepts

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

Algebraic Equations
Algebraic equations are foundational to understanding mathematics and are encountered in various fields. They consist of constants, variables, and operators forming an equality which we're expected to solve for the unknown variables.

For example, when we have an equation like \(2x - 4 = 10\), the goal is to determine the value of \(x\) that makes the equation true. Solving an algebraic equation involves performing operations that simplify the equation and isolate the variable, maintaining the equality by applying each operation to both sides.
Radical Expressions
Radical expressions contain a radical sign, which denotes the root of a number. The most common radical is the square root, but roots can be of any degree. When we work with radicals, we enter the realm of non-linear equations, which can introduce complexities in the solving process.

One key aspect to remember is that the square root of a number represents two values: one positive and one negative. However, when working with equations, we usually refer to the principal (positive) square root. For higher degree roots, the rules change slightly, especially for even and odd roots, impacting how we solve these equations.
No Real Solution
At times, equations may lead to a scenario where no real number can satisfy the given conditions. In the context of radical equations, this normally occurs when we have an even root equated to a negative number.

Since the square root of a real number can't produce a negative result, we conclude the equation has no real solution. This is important to recognize to avoid wasting time on unsolvable problems and to understand the limitations of real numbers in certain mathematical contexts.
Isolate the Radical
When solving radical equations, a key step is to 'isolate the radical.' This means that the term with the radical should be alone on one side of the equation. Doing so allows us to prepare the equation for the subsequent steps needed to find a solution.

Isolating the radical typically involves basic algebraic manipulation, such as adding, subtracting, multiplying, or dividing both sides of the equation by the same number or expression. This step is crucial for simplifying the equation and eventually eliminating the radical for further analysis.

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Most popular questions from this chapter

In Exercises \(75-92,\) rationalize each denominator. Simplify, if possible. $$\frac{5 \sqrt{3}-3 \sqrt{2}}{3 \sqrt{2}-2 \sqrt{3}}$$

In Exercises \(105-112,\) add or subtract as indicated. Begin by rationalizing denominators for all terms in which denominators contain radicals. $$\sqrt[4]{8}-\frac{20}{\sqrt[3]{2}}$$

The graph for Exercises \(55-56\) shows that the less income people have, the more likely they are to report fair or poor health. What explanations can you offer for this trend?

Use a graphing utility to solve each radical equation. Graph each side of the equation in the given viewing rectangle. The equation's solution set is given by the \(x\) -coordinate(s) of the point (s) of intersection. Check by substitution. $$\begin{aligned} &\sqrt{x}+3=5\\\ &[-1,6,1] \text { by }[-1,6,1] \end{aligned}$$

In Exercises \(75-92,\) rationalize each denominator. Simplify, if possible. $$\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}$$
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