/*! 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 420 Show that \(\mathrm{x}^{2}+2 \ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 420

Show that \(\mathrm{x}^{2}+2 \mathrm{x}+5=20\) is a conditional equation.

Short Answer

Expert verified
The equation \(x^2 + 2x + 5 = 20\) is a conditional equation because it holds true for two specific values of x: \(x_1 = 3\) and \(x_2 = -5\). We found these solutions using the quadratic formula and verified that the equation is true for both values.

Step by step solution

01

Simplify the equation.

First, let's simplify the equation by subtracting 20 from both sides. \(x^2 + 2x + 5 - 20 = 0\) The simplified equation will be: \(x^2 + 2x - 15 = 0\)
02

Solve the quadratic equation.

Here, we have a quadratic equation in the form of \(ax^2 + bx + c = 0\). To find the roots, we will determine two values of x for which the equation holds true. We can solve this quadratic equation using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\] In our equation, \(a = 1\), \(b = 2\), and \(c = -15\). Plugging these values into the quadratic formula, we get: \[x = \frac{-(2) \pm \sqrt{(2)^{2} - 4(1)(-15)}}{2(1)}\]
03

Simplify and calculate the roots.

Now, let's simplify the expression further: \[x = \frac{-2 \pm \sqrt{4 + 60}}{2}\] \[x = \frac{-2 \pm \sqrt{64}}{2}\] Since the square root of 64 is 8, we have: \[x = \frac{-2 \pm 8}{2}\] This gives us two possible solutions for x: \[x_1 = \frac{-2 + 8}{2} = 3\] and \[x_2 = \frac{-2 - 8}{2} = -5\]
04

Verify the solutions in the original equation.

We have found two solutions for x: \(x_1 = 3\) and \(x_2 = -5\). To show that the equation is conditional, we need to verify that the equation holds true for one or both of these values. We will substitute these values back into the original equation and check if the equation is true. For \(x_1 = 3\): \((3)^2 + 2(3) + 5 = 20\) \(9 + 6 + 5 = 20\) \(20 = 20\) (True) For \(x_2 = -5\): \((-5)^2 + 2(-5) + 5 = 20\) \(25 - 10 + 5 = 20\) \(20 = 20\) (True)
05

Conclusion.

Since the equation \(x^2 + 2x + 5 = 20\) holds true for \(x_1 = 3\) and \(x_2 = -5\), we can conclude that the equation is a conditional equation because it is true for these specific values of x.

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!

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

\(3 \mathrm{x}^{2}+5 \mathrm{x}=0\)

Find the roots of the equation \(x^{2}+6 x+8=0\)

Solve the equation \((3 \mathrm{x}-7)(\mathrm{x}+2)=0\)

Solve the equation \(\mathrm{x}^{2}+8 \mathrm{x}+15=0\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

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.