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Chapter 2: Problem 41

Solve the radical equation to find all real solutions. Check your solutions. $$\sqrt{x+1}+2=x$$

Short Answer

Expert verified
The real solution to the radical equation is \(x = 3\).

Step by step solution

01

Isolate the radical expression

Subtract 2 from both sides of the equation to make the square root term by itself on one side: \(\sqrt{x+1} = x-2\)
02

Square both sides

Square both sides of the equation to eliminate the square root. Remember the formula \((a-b)^2 = a^2 - 2ab + b^2\). This yields \(x+1 = (x-2)^2 = x^2 - 4x + 4\)
03

Simplify to quadratic equation

Transform the equation into a quadratic equation. To do it, bring all terms to one side: \(0 = x^2 - 4x - x + 4 - 1= x^2 - 5x + 3\)
04

Solve the quadratic equation

Solve this quadratic equation, by looking for factors of 3 that sum to 5. The factors are 1 and 3, hence the roots are 1 and 3, or \(x = 1, x = 3\).
05

Check the solutions

Check both solutions in the original equation. For \(x = 1\): \(\sqrt{1+1}+2 = 1+2 =3\) which is not equal to 1, so this is not a solution. For \(x = 3\): \(\sqrt{3+1}+2= 3\) which holds true, so 3 is a valid solution for the equation.

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Key Concepts

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

Square Root Equations
Solving square root equations, such as ewline
Quadratic Equations
Quadratic equations are in the form ewline
Isolation of Radical Expression
The isolation of a radical expression is a key strategy in solving equations that involve roots. Essentially, this means you want to get the term with the root by itself on one side of the equation to facilitate the removal of the root. This isolation step is crucial for the subsequent steps in solving radical equations because it prepares the equation for squaring, which eliminates the radical. For example, in the step by step solution provided, the first thing we did was move the 2 from the left-hand side to the right-hand side to isolate the ewline
Checking Solutions in Algebra
When solving algebraic equations, checking your solutions in the context of the original equation is crucial to verify their correctness. This is especially important when dealing with radical equations because the act of squaring both sides can introduce extraneous solutions. These are solutions that appear valid when plugged into the modified equation but do not satisfy the original equation. After identifying potential solutions, you should substitute them back into the original radical equation to confirm that they produce a true statement.

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Most popular questions from this chapter

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=\frac{-x+1}{2 x+3} ; g(x)=\frac{1}{x^{2}+1}$$

In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=3 x-1$$

Graph each quadratic function by finding a suitable viewing window with the help of the TABLE feature of a graphing utility. Also find the vertex of the associated parabola using the graphing utility. $$s(t)=-16 t^{2}+40 t+120$$

This set of exercises will draw on the ideas presented in this section and your general math background. Why must we have \(a \neq 0\) in the definition of a quadratic function?

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=\sqrt[5]{-x^{3}+8}$$
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