/*! 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 27 Solve the inequality by factorin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the inequality by factoring. $$-x^{2}+2 x-1<0$$

Short Answer

Expert verified
The inequality \(-x^{2}+2 x-1<0\) has no solution.

Step by step solution

01

Factorize the quadratic polynomial

Firstly, the quadratic polynomial \(-x^{2}+2 x-1\) is factored to find its roots. It is factored into \(-(x-1)^2\).
02

Find the critical points

The roots or zeros of the equation represent the critical points. Set \(-(x-1)^2 = 0\). The solution for x is \(1\). Thus, \(1\) is the critical point.
03

Determine the intervals

The critical point splits the number line into two intervals: \(-\infty ,1\) and \(1,+\infty \). These intervals will be tested to determine where the function is negative and satisfies the inequality.
04

Test the intervals

Test a value from each interval in the factored equation \(-(x-1)^2\), here, for example 0 and 2 can be used, for \(-\infty ,1\) and \(1,+\infty \) respectively. Substituting these values in, for \(x=0\), \((0-1)^2=1>0\) and for \(x=2\), \((2-1)^2=1>0\). For both intervals, the result is positive, therefore no interval satisfies the inequality.
05

Answer

Based upon the tests, there are no intervals in which the inequality \(-x^{2}+2 x-1<0\) holds true.

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.

Factoring Quadratics
Factoring quadratics is an essential technique in algebra and helps to simplify quadratic expressions by representing them as the product of their factors. These factors can then be used to find the roots or solutions of the quadratic equation. In the given inequality \(-x^{2}+2x-1<0\), factoring is applied initially to help solve the inequality. This expression can be rewritten as \(- (x-1)^2 \), by identifying \(x-1\) as the repeated factor for the original polynomial. By successfully factoring, we begin to see the equation in its simplest terms.
Understanding how to factor is crucial because it transforms complex expressions into more manageable forms which are easier to work with.
  • Identify any common factors that all terms share.
  • Recognize patterns such as perfect square trinomials or the difference of squares.
  • Rearrange terms and apply the appropriate factoring operations.
Critical Points
Critical points occur where the expression being evaluated changes its sign or its slope becomes zero. These points are often the solutions or roots of the quadratic where it equals zero. In our inequality, setting the factored quadratic \(-(x-1)^2=0\) reveals that the equation only has the root \(x = 1\). This root is our critical point.
The critical point divides the number line into distinct intervals where the value of the quadratic will behave differently. Analyzing behavior at the critical points helps understand how the function changes across the number line, assisting in solving inequalities.
  • Found by setting the factored form equal to zero.
  • Helps in breaking down intervals to test for sign changes in the inequality.
  • Critical for determining where inequalities hold true.
Interval Testing
Interval testing involves plugging different values from specific intervals back into the factored equation to determine whether they satisfy the inequality. Since our inequality is separated into intervals by \(x = 1\), we have two main intervals to test: \(-\infty, 1\) and \(1, \infty\). By substituting values like 0 from \(-\infty, 1\) and 2 from \(1, \infty\),we can observe how the inequality behaves.
For instance, at \(x = 0\), \(-(0-1)^2\) results in a positive outcome, indicating it does not satisfy the inequality \(-x^2+2x-1 < 0\). Moreover, when checking \(x = 2\), \(-(2-1)^2\) also results in a positive number, further confirming this behavior for the whole interval.
  • Exercises involve choosing test values well within the interval.
  • Each test value will determine if the inequality is satisfied.
  • Testing both sides of a critical point helps confirm solutions or lack thereof.
Inequality Solving Methods
Solving inequalities using factoring and interval testing requires a grasp of various techniques that extend beyond simply solving for \(x\). We navigate quadratic inequalities by transforming them into a product of factors, finding critical points, and testing intervals. Here, methods such as these are combined to deduce where the quadratic expression is less than zero.
Ultimately, when all interval tests resulted in a positive inequality in our specific exercise, it displays how no value of \(x\) actually satisfies this inequality \(-x^2 + 2x - 1 < 0\). Other common methods involve graphing or completing the square, but in this example, the emphasis remains on algebraic techniques.
  • Combines factoring, interval testing, and critical points for comprehensive understanding.
  • Crucial for understanding where solutions occur in inequalities.
  • Serves as a handy toolset to approach various forms of inequalities efficiently.

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

The production cost, in dollars, for \(x\) color brochures is \(C(x)=500+3 x .\) The fixed cost is \(\$ 500\) since that is the amount of money needed to start production even if no brochures are printed. (a) If the fixed cost is decreased by \(\$ 50,\) find the new cost function. (b) Graph both cost functions and interpret the effect of the decreased fixed cost.

A rectangular fence is being constructed around a new play area at a local elementary school. If the school has 2000 feet of fencing available for the project, what is the maximum area that can be enclosed for the new play area?

This set of exercises will draw on the ideas presented in this section and your general math background. Without doing any calculations, explain why $$\sqrt{x+1}=-2$$ does not have a solution.

A parabola associated with a certain quadratic function \(f\) has the point (2,8) as its vertex and passes through the point \((4,0) .\) Find an expression for \(f(x)\) in the form \(f(x)=a(x-h)^{2}+k\) (a) From the given information, find the values of \(h\) and \(k\) (b) Substitute the values you found for \(h\) and \(k\) into the expression \(f(x)=a(x-h)^{2}+k\) (c) Now find \(a\). To do this, use the fact that the parabola passes through the point \((4,0) .\) That is, \(f(4)=0\) You should get an equation having just \(a\) as a variable. Solve for \(a\) (d) Substitute the value you found for \(a\) into the expression you found in part (b). (e) Graph the function using a graphing utility and check your answer. Is (2,8) the vertex of the parabola? Does the parabola pass through (4,0)\(?\)

In Exercises \(97-100,\) let \(f(t)=-t^{2}\) and \(g(x)=x^{2}-1\). Find an expression for \((g \circ g)(x),\) and give the domain of \(g \circ g\).

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.