/*! 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 49 Solve each equation. Check your ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 49

Solve each equation. Check your solutions. $$ x^{4}-29 x^{2}+100=0 $$

Short Answer

Expert verified
The solutions are \( x = \pm 5 \) and \( x = \pm 2 \).

Step by step solution

01

- Make a Substitution

Let's start by making a substitution to simplify the equation. Let鈥檚 set \( u = x^2 \). Therefore, the original equation becomes \( u^2 - 29u + 100 = 0 \).
02

- Solve the Quadratic Equation

Now, solve the quadratic equation \( u^2 - 29u + 100 = 0 \). To factor this, we look for two numbers that multiply to 100 and add up to -29. These numbers are -25 and -4. Therefore, the equation factors as \( (u - 25)(u - 4) = 0 \).
03

- Solve for u

Now set each factor equal to zero and solve for \( u \): \( u - 25 = 0 \) gives \( u = 25 \), and \( u - 4 = 0 \) gives \( u = 4 \).
04

- Substitute Back for x

Recall that \( u = x^2 \). So, substitute back to get \( x^2 = 25 \) and \( x^2 = 4 \).
05

- Solve for x

Solve these equations for \( x \): \( x^2 = 25 \) gives \( x = \pm 5 \), and \( x^2 = 4 \) gives \( x = \pm 2 \).
06

- Check Solutions

Check these solutions by substituting back into the original equation: \( x = 5, -5, 2, -2 \). For \( x = 5 \), \( 5^4 - 29 \times 5^2 + 100 = 0 \), for \( x = -5 \), \( (-5)^4 - 29 \times (-5)^2 + 100 = 0 \), for \( x = 2 \), \( 2^4 - 29 \times 2^2 + 100 = 0 \), and for \( x = -2 \), \( (-2)^4 - 29 \times (-2)^2 + 100 = 0 \). All checks confirm the solutions.

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 91影视!

Key Concepts

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

quadratic substitution
To simplify the process of solving polynomial equations, especially higher-degree ones, we can use quadratic substitution. This method helps convert a complex polynomial into a simpler quadratic equation that we can easily work with. In our exercise, we began with the equation: \( x^4 - 29x^2 + 100 = 0 \). By setting \( u = x^2 \), we transformed it into a quadratic form: \( u^2 - 29u + 100 = 0 \). This substitution reduces the complexity of our original equation, providing an easier pathway to solution.
factoring quadratic equations
Once we have a quadratic equation like \( u^2 - 29u + 100 = 0 \), we can solve it by factoring. Factoring involves finding two numbers that multiply to the constant term (100) and add up to the linear coefficient (-29). In this case, the suitable numbers are -25 and -4. By expressing the equation in factored form, we get:\( (u - 25)(u - 4) = 0 \). Now we can set each factor equal to zero and solve for \( u \): \( u - 25 = 0 \) gives \( u = 25 \), and \( u - 4 = 0 \) gives \( u = 4 \).
solving higher-degree polynomials
After solving the quadratic equation, we need to return to our original variable \( x \). Recalling that \( u = x^2 \), we substitute back and solve for \( x \). The values of \( u \) that we obtained are 25 and 4, so we set up the equations: \( x^2 = 25 \) and \( x^2 = 4 \). By solving these, we find:- \( x = \,pm 5 \).- \( x = \,pm 2 \).Therefore, the solutions to the original polynomial equation are \( x = 5, -5, 2, \,and -2 \). Verifying these solutions by substituting them back into the original equation confirms their correctness.

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

Match each quadratic function with the description of the parabola that is its graph. (a) \(f(x)=(x-4)^{2}-2\) (b) \(f(x)=(x-2)^{2}-4\) (c) \(f(x)=-(x-4)^{2}-2\) (d) \(f(x)=-(x-2)^{2}-4\) A. Vertex \((2,-4),\) opens down B. Vertex \((2,-4),\) opens up C. Vertex \((4,-2),\) opens down D. Vertex \((4,-2),\) opens up

Solve for \(x .\) Assume that a and b represent positive real numbers. \(x^{2}-b=0\)

Solve each equation for the indicated variable. (Leave \(\pm\) in your answers.) $$ S=4 \pi r^{2} \text { for } r $$

Solve each equation for the indicated variable. (Leave \(\pm\) in your answers.) $$ S=2 \pi r h+\pi r^{2} \text { for } r $$

Find the nonreal complex solutions of each equation. \(4 x^{2}+5 x+5=0\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation
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.