/*! 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 62 In solving \(\sqrt{2 x-1}+2=x,\)... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 62

In solving \(\sqrt{2 x-1}+2=x,\) why is it a good idea to isolate the radical term? What if we don't do this and simply square each side? Describe what happens.

Short Answer

Expert verified
The solution to the equation is \(x = 1\). When solving equations with radicals, it's important to isolate the radical before squaring both sides. If not, extraneous solutions might arise which further need to be checked for validity.

Step by step solution

01

Isolate the radical

The first step is to isolate the radical part of the equation from the rest. This involves moving the non-radical terms to the other side of the equation. To do this, subtract 2 from both sides for the equation: \(\sqrt{2x - 1} = x - 2\)
02

Remove the radical

Second, to remove the radical, square both sides of the equation. But note that here arises the problem of extraneous solutions which need to be checked before deciding the final solution (an extraneous solution is a solution that appears valid when plugged into the original equation but is not a valid solution for the original equation). Squaring both sides yields: \((\sqrt{2x - 1})^2 = (x - 2)^2\), which simplifies to \(2x - 1 = x^2 - 4x + 4\)
03

Simplify the equation

Now, simplify the equation into a standard quadratic form by moving all terms to one side of the equation, thus the equation becomes: \(x^2 - 6x + 5 = 0\)
04

Solve the quadratic equation

The next step involves solving the quadratic equation for \(x\). This can be done by factoring or using the quadratic formula. The quadratic equation factors as: \((x - 1)(x - 5) = 0\). Setting each factor equal to zero gives us \(x = 1\) and \(x = 5\)
05

Check for extraneous solutions

Check if these solutions are extraneous by plugging them back into the original equation. After substitution one can observe that only \(x = 1\) remains valid, thus \(x = 5\) is an extraneous solution.

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Most popular questions from this chapter

Simplify each expression. Write answers in exponential form with positive exponents only. Assume that all variables represent positive real numbers. $$x^{\frac{1}{4}} \cdot x^{\frac{1}{5}}$$

Simplify by first writing the expression in radical form. If applicable, use a calculator to verify your answer. $$32^{-\frac{4}{5}}$$

Do you expect to pay more taxes than were withheld? Would you be surprised to know that the percentage of taxpayers who receive a refund and the percentage of taxpayers who pay more taxes vary according to age? The formula $$ P=\frac{x(13+\sqrt{x})}{5 \sqrt{x}} $$ models the percentage, \(P\), of taxpayers who are \(x\) years old who must pay more taxes. a. What percentage of 25 -year-olds must pay more taxes? b. Rewrite the formula by rationalizing the denominator. c. Use the rationalized form of the formula from part (b) to find the percentage of 25 -year-olds who must pay more taxes. Do you get the same answer as you did in part (a)? If so, does this prove that you correctly rationalized the denominator? Explain.

Simplify by first writing the expression in radical form. If applicable, use a calculator to verify your answer. $$243^{-\frac{1}{5}}$$

Describe what it means to rationalize a denominator. Use both \(\frac{1}{\sqrt{5}}\) and \(\frac{1}{5+\sqrt{5}}\) in your explanation.
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