/*! 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 36 Solve the equation by using the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 36

Solve the equation by using the quadratic formula where appropriate. $$5 u^{2}=3$$

Short Answer

Expert verified
u = \( \frac{\pm \sqrt{15}}{5} \)

Step by step solution

01

Write the equation in standard quadratic form

Given the equation: \[5u^2 = 3\] To use the quadratic formula, the equation must be in the form \[au^2 + bu + c = 0\]. Subtract 3 from both sides to obtain: \[5u^2 - 3 = 0\]
02

Identify coefficients

Compare the equation \[5u^2 - 3 = 0\] with \[au^2 + bu + c = 0\]. Here, we have: \(a = 5\), \(b = 0\), and \(c = -3\)
03

Apply the quadratic formula

The quadratic formula is: \[u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substitute the values of \(a\), \(b\), and \(c\): \[u = \frac{-(0) \pm \sqrt{(0)^2 - 4(5)(-3)}}{2(5)}\]
04

Simplify under the square root

Simplify the expression under the square root: \[u = \frac{0 \pm \sqrt{0 + 60}}{10} = \frac{\pm \sqrt{60}}{10}\]
05

Simplify the square root

The value \(\sqrt{60}\) can be simplified to \(\sqrt{4 \times 15} = 2\sqrt{15}\). Substituting this back gives: \[u = \frac{\pm 2\sqrt{15}}{10} = \frac{\pm \sqrt{15}}{5}\]

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.

Solving Quadratic Equations
Solving quadratic equations often involves finding the values of a variable that make the equation true. Quadratic equations typically look like \(au^2 + bu + c = 0\). In this exercise, we used the quadratic formula to find the roots or solutions for the equation \[5u^2 = 3\]. To do this, we first wrote it in the standard form \(au^2 + bu + c = 0\). We then identified the coefficients and plugged them into the quadratic formula, which is \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Following these steps methodically will help you find the solutions of any quadratic equation.
Standard Quadratic Form
The standard quadratic form is \(au^2 + bu + c = 0\). This is important because the quadratic formula can only be applied to equations in this form. For example, our given equation \[5u^2 = 3\] wasn't in standard form initially. To convert it, we subtracted 3 from both sides to get \(5u^2 - 3 = 0\). Now, it matches \(au^2 + bu + c = 0\), where \(a = 5\), \(b = 0\), and \(c = -3\). From here, it鈥檚 easier to use the quadratic formula to solve for \(u\).
Simplifying Square Roots
Simplifying square roots is a crucial skill for solving quadratic equations. When we applied the quadratic formula in this exercise, we encountered the square root \(\sqrt{60}\). Simplifying square roots involves finding factor pairs, one of which is a perfect square. For example, \(60 = 4 \times 15\). The perfect square here is 4. Thus, \(\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}\). This simplification helps keep the solutions in their simplest form, making them easier to understand and use. Remember to always look for perfect square factors to simplify square roots 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

In Exercises \(1-64\), solve each of the given equations. If the equation is quadratic, use the factoring or square root method. If the equation has no real solutions, say so. $$a+\frac{3 a}{a-3}=\frac{9}{a-3}$$

In Exercises \(1-64\), solve each of the given equations. If the equation is quadratic, use the factoring or square root method. If the equation has no real solutions, say so. $$\frac{3}{x-2}+\frac{7}{x+2}=\frac{x+1}{x-2}$$

In Exercises \(75-84\), round your answer to the nearest tenth where necessary. The legs of a right triangle are \(15 \mathrm{mm}\) and \(20 \mathrm{mm}\). Find the length of the hypotenuse.

Use the method you think is the most appropriate to solve the given equation. Check your answers by using a different method. $$x^{2}+\frac{2}{3} x=4$$

In Exercises \(1-64\), solve each of the given equations. If the equation is quadratic, use the factoring or square root method. If the equation has no real solutions, say so. $$\left(x+\frac{2}{5}\right)^{2}=\frac{3}{25}$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete 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.