/*! 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 38 Solve each equation, and check t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 38

Solve each equation, and check the solutions. $$ 6 r^{2}-r-2=0 $$

Short Answer

Expert verified
\(r = \frac{2}{3}\) and \(r = -\frac{1}{2}\)

Step by step solution

01

- Identify the Standard Form

The given equation is already in the standard quadratic form: \[6r^2 - r - 2 = 0\]
02

- Apply the Quadratic Formula

For the quadratic equation \[ax^2 + bx + c = 0\], use the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 6\), \(b = -1\), and \(c = -2\).
03

- Calculate the Discriminant

Calculate the discriminant \(\Delta = b^2 - 4ac\):\[\Delta = (-1)^2 - 4(6)(-2) = 1 + 48 = 49\]
04

- Find the Roots Using the Discriminant

Substitute \(\Delta\) back into the quadratic formula: \[r = \frac{-(-1) \pm \sqrt{49}}{2 \times 6} = \frac{1 \pm 7}{12}\]This results in two solutions: \[r = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}\]\[r = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}\]
05

- Check the Solutions

Substitute \(r = \frac{2}{3}\) and \(r = -\frac{1}{2}\) back into the original equation to verify:For \(r = \frac{2}{3}\):\[6 \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right) - 2 = 6 \left(\frac{4}{9}\right) - \frac{2}{3} - 2 = \frac{24}{9} - \frac{2}{3} - 2 = \frac{8}{3} - \frac{2}{3} - 2 = \frac{6}{3} - 2 = 0\]For \(r = -\frac{1}{2}\):\[6 \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) - 2 = 6 \left(\frac{1}{4}\right) + \frac{1}{2} - 2 = \frac{6}{4} + \frac{1}{2} - 2 = \frac{3}{2} + \frac{1}{2} - 2 = 2 - 2 = 0\]Both solutions satisfy the original equation.

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 Formula
The quadratic formula is a powerful tool for solving quadratic equations. A quadratic equation is an equation in the form ewline ax^2 + bx + c = 0, where a, b, and c are constants, and x is the variable. To solve for x, we use the quadratic formula: ewline \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] ewline ewline Simply plug in the values of a, b, and c from your equation into this formula. You鈥檒l get two solutions which are the values of x that satisfy the equation. This is because of the \pm symbol, indicating there will be two values: one for the plus and one for the minus.
Discriminant
The discriminant is part of the quadratic formula, specifically the term under the square root: ewline \[ \Delta = b^2 - 4ac \] ewline ewline The discriminant tells us a lot about the nature of the roots of the quadratic equation: ewline
  • If \Delta > 0, there are two distinct real roots.
  • If \Delta = 0, there is exactly one real root (the roots are repeated).
  • If \Delta < 0, there are no real roots (the solutions are complex).
ewline ewline In our example ewline \[6r^2 - r - 2 = 0 \, a = 6, b = -1, \text{ and } c = -2 \] ewline ewline Calculating the discriminant gives: ewline \[ \Delta = (-1)^2 - 4(6)(-2) = 1 + 48 = 49 \] ewline Since \Delta = 49 which is positive, we know there are two distinct real roots for this quadratic equation.
Standard Quadratic Form
The standard quadratic form is how we usually express quadratic equations: ewline \[ ax^2 + bx + c = 0 \] ewline Notice each part where:
  • a is the coefficient of the quadratic term ( x^2 ).
  • b is the coefficient of the linear term (x).
  • c is the constant term.
ewline ewline To solve any quadratic equation, it鈥檚 important to first rearrange the equation into this format. In our exercise: ewline \[6r^2 - r - 2 = 0\] ewline ewline It鈥檚 already in the standard form with a = 6, b = -1, and c = -2. Once in standard form, you can apply the quadratic formula to find the solutions easily.

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

Solve each problem. The length of a rectangle is 5 in. longer than its width. The diagonal is 5 in. shorter than twice the width. Find the length, width, and diagonal measures of the rectangle.

Factor each binomial completely. \(125 m^{3}+8 p^{3}\)

Solve each problem. The product of the page numbers on two facing pages of a book is 600 . Find the page numbers.

Factor each binomial completely. \(64 x^{3}+125 y^{3}\)

The following powers of \(x\) are all perfect cubes: $$ x^{3}, x^{6}, \quad x^{9}, \quad x^{12}, \quad x^{15} $$ On the basis of this observation, we may make a conjecture that if the power of a variable is divisible by ____ (with 0 remainder), then we have a perfect cube.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

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.