/*! 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 83 Find the zeros of the function a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 83

Find the zeros of the function algebraically. Give exact answers. $$f(x)=5 x^{2}-2 x-1$$

Short Answer

Expert verified
The zeros of the function are \(x = \frac{1 + \sqrt{6}}{5}\) and \(x = \frac{1 - \sqrt{6}}{5}\)

Step by step solution

01

Identify the quadratic equation

The function given is a quadratic equation in the form of \[f(x) = 5x^{2} - 2x - 1\]
02

Write the quadratic equation in standard form

The standard form of a quadratic equation is given by \[ax^{2} + bx + c = 0\]. For the given function, this is \[5x^{2} - 2x - 1 = 0\]
03

Apply the quadratic formula

Use the quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Here, \(a = 5\), \(b = -2\), \(c = -1\)
04

Calculate the discriminant

The discriminant of the quadratic formula is \[b^2 - 4ac\]. Substitute \(a = 5\), \(b = -2\), \(c = -1\): \[(-2)^2 - 4(5)(-1) = 4 + 20 = 24\]
05

Solve for x using the quadratic formula

Substitute the values and the discriminant into the quadratic formula: \[x = \frac{-(-2) \pm \sqrt{24}}{2(5)}\]. Simplify to get the solutions: \[x = \frac{2 \pm \sqrt{24}}{10} = \frac{2 \pm 2\sqrt{6}}{10} = \frac{1 \pm \sqrt{6}}{5}\]

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 Equations
A quadratic equation is a type of polynomial equation of the form \[ax^2 + bx + c = 0\], where \a\, \b\, and \c\ are constants and \a eq 0\. Quadratic equations are called 'quadratic' because \'quad\' means square, and their highest exponent is 2, meaning the solution involves squared terms. Understanding quadratic equations is fundamental because they appear in various contexts like physics, engineering, finance, and more.
For our exercise, the given quadratic equation was \[5x^2 - 2x - 1 = 0\]. Here, \a = 5\, \b = -2\, and \c = -1\. Solving this equation involves methods like factoring, completing the square, or more commonly using the quadratic formula.
Quadratic Formula
The quadratic formula is a powerful tool used to solve any quadratic equation. The formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula provides the exact solutions (zeros) of the quadratic equation \[ax^2 + bx + c = 0\] and is derived by completing the square on the general quadratic equation.
To use the quadratic formula:
  • Identify the coefficients \a\, \b\, and \c\ from the given equation.
  • Calculate the discriminant, \b^2 - 4ac\.
  • Substitute \a\, \b\, and the discriminant values into the quadratic formula.
In our example, we found values \a = 5\, \b = -2\, and \c = -1\.
Discriminant Calculation
The discriminant, represented by \[b^2 - 4ac\], helps determine the nature of the solutions of a quadratic equation without actually solving it:
  • If \b^2 - 4ac > 0\, there are two distinct real solutions.
  • If \b^2 - 4ac = 0\, there is one real solution (a repeated or double root).
  • If \b^2 - 4ac < 0\, there are no real solutions, only complex ones.
For our given quadratic equation \[5x^2 - 2x - 1 = 0\], we computed the discriminant as follows:
\[(-2)^2 - 4(5)(-1) = 4 + 20 = 24\]
Since the discriminant is positive (24 > 0), the equation has two distinct real solutions, which were calculated using the quadratic formula.

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 stone is thrown directly upward from a height of 30 ft with an initial velocity of \(60 \mathrm{ft} / \mathrm{sec} .\) The height of the stone, in feet, \(t\) seconds after it has been thrown is given by the function \(s(t)=-16 t^{2}+60 t+30 .\) Determine the time at which the stone reaches its maximum height and find the maximum height.

Solve. \(\frac{V^{2}}{R^{2}}=\frac{2 g}{R+h},\) for \(h\) (A formula for escape velocity)

a) Find the vertex. b) Determine whether there is a maximum or a minimum value and find that value. c) Find the range. d) Find the intervals on which the function is increasing and the intervals on which the function is decreasing. $$g(x)=2 x^{2}-6 x+5$$

Solve and write interval notation for the solution set. Then graph the solution set. $$|4 x|>20$$

Solve. $$\sqrt{y+4}-\sqrt{y-1}=1$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

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.