/*! 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 30 Solve each equation. \(2 x-\sq... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each equation. \(2 x-\sqrt{x}-1=0\)

Short Answer

Expert verified
The solution is \( x = 1 \).

Step by step solution

01

Isolate the Square Root Term

Start by isolating the square root term on one side of the equation. The equation is initially given as: \[ 2x - \sqrt{x} - 1 = 0 \]Add \( \sqrt{x} + 1 \) to both sides to isolate the square root term:\[ 2x - 1 = \sqrt{x} \]
02

Square Both Sides

Square both sides of the equation to eliminate the square root. When you square both sides, remember to apply the square to each term:\[ (2x - 1)^2 = (\sqrt{x})^2 \]This simplifies to:\[ 4x^2 - 4x + 1 = x \]
03

Move All Terms to One Side

Move all the terms to one side of the equation to form a quadratic equation:\[ 4x^2 - 4x + 1 - x = 0 \]Simplify and write it as:\[ 4x^2 - 5x + 1 = 0 \]
04

Solve the Quadratic Equation

This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \). Use the quadratic formula to solve for \( x \):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 4 \), \( b = -5 \), and \( c = 1 \). Plug these values into the formula:\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]Simplify inside the square root first:\[ x = \frac{5 \pm \sqrt{25 - 16}}{8} \]\[ x = \frac{5 \pm \sqrt{9}}{8} \]\[ x = \frac{5 \pm 3}{8} \]
05

Find the Potential Solutions

Solve for the two potential solutions using the simplified quadratic result:\[ x_1 = \frac{5 + 3}{8} = 1 \]\[ x_2 = \frac{5 - 3}{8} = \frac{1}{4} \]
06

Check Both Solutions in the Original Equation

For potential solution \( x_1 = 1 \):\[ 2(1) - \sqrt{1} - 1 = 0 \] simplifies to \( 2 - 1 - 1 = 0 \), which is true.For potential solution \( x_2 = \frac{1}{4} \):\[ 2\left(\frac{1}{4}\right) - \sqrt{\frac{1}{4}} - 1 = 0 \]\[ \frac{1}{2} - \frac{1}{2} - 1 eq 0 \], which is false. Thus, \( x_2 \) is not a solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Square Root Isolation
In many quadratic equations, especially those involving square roots, the first step is isolating the square root term. This makes it simpler to deal with and allows the rest of the equation to be manipulated more easily.

When beginning with the equation \(2x - \sqrt{x} - 1 = 0\), your goal is to get \(\sqrt{x}\) by itself on one side of the equation. To do this, you add \(\sqrt{x} + 1\) to both sides of the equation, which gives you:
  • \(2x - 1 = \sqrt{x}\)
This is an essential step, as it separates the square root so you can eliminate it through squaring. After isolation, you manage to transform the problem into one where you can apply algebraic manipulation like squaring both sides to further simplify and solve the equation. Remember, isolating square roots is crucial to correctly handling the problem and will set you on a path to finding possible solutions.
Quadratic Formula
Once the square root has been isolated and the equation has been squared, you often end up with a quadratic equation, which is typical after squaring both sides. In this instance, the equation becomes \(4x^2 - 4x + 1 = x\). After rearranging—as shown by moving \(x\) to one side—you get the classic form:
  • \(4x^2 - 5x + 1 = 0\)
This is now ready for the powerful quadratic formula:
  • \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our quadratic equation, we have \(a = 4\), \(b = -5\), and \(c = 1\). By substituting these into the formula, we can find the potential solutions:
  • \(x = \frac{5 \pm \sqrt{9}}{8}\), which simplifies to \(x = 1\) and \(x = \frac{1}{4}\)
Using the quadratic formula lets us solve for \(x\) efficiently, particularly when the quadratic cannot be easily factored. This handy formula is a go-to method for solving such equations.
Verifying Solutions
After solving any quadratic equation, verifying your solutions is a necessary step to ensure accurate results. This is because not all solutions derived might satisfy the original equation, especially when initial equations involve square roots. For the solutions derived as \(x = 1\) and \(x = \frac{1}{4}\), you insert them back into the initial equation \(2x - \sqrt{x} - 1 = 0\):
  • For \(x = 1\), the equation simplifies perfectly to \(2 - 1 - 1 = 0\), which is correct.
  • For \(x = \frac{1}{4}\), it results in \(\frac{1}{2} - \frac{1}{2} - 1 eq 0\), clearly indicating that it does not satisfy the original equation.
Verifying is crucial because earlier steps like squaring both sides can introduce extraneous solutions. Confirming which solutions hold true maintains the integrity of the mathematical process and ensures understanding of the methodology.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A golden rectangle is one of the most visually appealing of all geometric forms. The Parthenon, built by the Greeks in the 5 th century B.C., fits into a golden rectangle if its ruined triangular pediment is included. See the illustration. In a golden rectangle, the length \(l\) and width \(w\) must satisfy the equation \(\frac{l}{w}=\frac{w}{l-w} .\) If a rectangular billboard is to have a width of 20 feet, what should its length be so that it is a golden rectangle? Round to the nearest tenth.

Use a graphing calculator to solve each equation. If an answer is not exact, round to the nearest hundredth. See Using Your Calculator: Solving Quadratic Equations Graphically. Fireworks. \(\quad\) A fireworks shell is shot straight up with an initial velocity of 120 feet per second. Its height \(s\) in feet after \(t\) seconds is approximated by the equation \(s=120 t-16 t^{2} .\) If the shell is designed to explode when it reaches its maximum height, how long after being fired, and at what height, will the fireworks appear in the sky?

Operating costs. The cost \(C\) in dollars of operating a certain concrete- cutting machine is related to the number of minutes \(n\) the machine is run by the function \(C(n)=2.2 n^{2}-66 n+655\) For what number of minutes is the cost of running the machine a minimum? What is the minimum cost?

What are some quantities that are good to maximize? What are some quantities that are good to minimize?

Simplify each expression. Assume all variables represent positive numbers. $$ -2 \sqrt{5 b}(4 \sqrt{2 b}-3 \sqrt{3}) $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.