/*! 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 10: 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
If the square root term is not isolated before squaring, the resulting equation becomes far more complicated to solve. Therefore, always isolate the square root before squaring. In the given question, the correct answer turns out to be \( x=1 \).

Step by step solution

01

Isolating square root and squaring

Firstly, isolate the square root term on one side before squaring. The given equation is \( \sqrt{2x - 1} + 2 = x \). Isolate the square root term to look like this: \( \sqrt{2x - 1} = x - 2 \). Now, square both sides. This yields: \(2x - 1 = (x - 2)^2\).
02

Expansion and simplification

Expand the square term on the right side and simplify by placing all terms on the left side. This results in \(2x - 1 = x^2 - 4x + 4\), which simplifies to \(x^2 - 6x + 5 = 0\).
03

Solving the quadratic equation

Solve the resulting quadratic equation either by factoring, completing the square, or using the quadratic formula. In this case, the equation can be factored easily into this form: \( (x-5)(x-1) = 0 \). Therefore, the solutions are \( x=1 \) and \( x=5 \).
04

Check solution in original equation

Check these solutions by substituting them in the original equation. If a number makes the original equation true, then it is a valid solution. However, \( x=5 \) simply does not satisfy the original equation. Thus, the only solution to the original equation is \( x=1 \).
05

Procedure without isolating the square root

What if the square root term weren't isolated before squaring? If both sides of the initial equation were squared without isolating the square root, the result would be: \(2x - 1 + 4\sqrt{2x - 1} + 4 = x^2\). This leads to a much more complicated equation involving a square root term that is not isolated, making the equation harder to solve.

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Most popular questions from this chapter

Exercises \(145-147\) show a number of simplifications, not all of which are correct. Enter the left side of each equation as \(y_{1}\) and the right side as \(y_{2} .\) Then use your graphing utility's \(|\) TABLE feature to determine if the simplification is correct. If it is not, correct the right side and use the \([\text { TABLE }]\) feature to verify your simplification. $$\left(x^{-\frac{1}{2}} \cdot x^{\frac{3}{4}}\right)^{-2}=x^{\frac{1}{2}}$$

evaluate each expression, or state that the expression is not a real number. $$-\sqrt{0.64}$$

When a radical expression has its denominator rationalized, we change the denominator so that it no longer contains any radicals. Doesn't this change the value of the radical expression? Explain.

Explain how to solve a radical equation with rational exponents.

In Exercises \(85-100,\) simplify each expression. $$i^{22}$$
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