/*! 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 each equation. $$ 3 x^... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 36

Solve each equation. $$ 3 x^{2}+9 x+30=2 x^{2}-2 x $$

Short Answer

Expert verified
x = -5 or x = -6

Step by step solution

01

- Move all terms to one side

Subtract \[2x^{2} - 2x\] from both sides to move all terms to one side of the equation: \[3x^{2} + 9x + 30 - 2x^{2} + 2x = 0\]
02

- Combine like terms

Combine the like terms on the left side: \[x^{2} + 11x + 30 = 0\]
03

- Factor the quadratic equation

Factor the quadratic equation \[x^{2} + 11x + 30\]. Look for two numbers that multiply to 30 and add up to 11. These numbers are 5 and 6: \[(x + 5)(x + 6) = 0\]
04

- Solve for x

Set each factor equal to zero and solve for \x\: \[x + 5 = 0\] or \[x + 6 = 0\] \x = -5\ or \x = -6\

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 polynomial equation of the second degree. This means it includes an x虏 term, and its general form can be written as \(ax^{2} + bx + c = 0\).
These equations can have up to two real solutions, depending on the discriminant value \(b^{2} - 4ac\).
In our exercise, the quadratic equation we are solving is \(3x^{2} + 9x + 30 = 2x^{2} - 2x\).
To solve quadratic equations, we often use methods like factoring, using the quadratic formula, or completing the square.
factoring
Factoring is one of the most common methods to solve quadratic equations. By factoring, we write the quadratic equation as a product of two binomial expressions.
In the given exercise, we first rearrange the equation to combine like terms: \(x^{2} + 11x + 30 = 0\).
Next, we need to identify two numbers that multiply to give the constant term (30) and add up to give the coefficient of the middle term (11).
In this case, those numbers are 5 and 6. So we factor the quadratic equation into \((x + 5)(x + 6)\).
Factoring converts a quadratic equation into a simpler form, making it easier to solve for the variable x.
combining like terms
Combining like terms is a crucial step in simplifying and solving equations. Like terms are terms that have the same variable raised to the same power.
In the provided exercise, after subtracting \(2x^{2} - 2x\) from both sides, we get: \(3x^{2} + 9x + 30 - 2x^{2} + 2x = 0\).
We then combine the like terms: \(x^{2} + 11x + 30 = 0\).
This step simplifies the equation to a standard quadratic form, which makes it easier to factor and eventually solve for x.
solving for x
Solving for x is the final step after simplifying and factoring the quadratic equation. Once we have the equation in the factored form, \((x + 5)(x + 6) = 0\), we set each factor equal to zero: \(x + 5 = 0\) and \(x + 6 = 0\).
This gives us the solutions \(x = -5\) and \(x = -6\).
By solving these simple linear equations, we find the values of x that satisfy the original quadratic equation. This method ensures that all possible solutions are captured.

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

The binomial \(x^{6}-y^{6}\) may be considered as either a difference of squares or a difference of cubes. Factor \(x^{6}-y^{6}\) by first factoring as a difference of squares. Then factor further by considering one of the factors as a sum of cubes and the other factor as a difference of cubes.

Solve each equation. $$ (x+3)(x-6)=(2 x+2)(x-6) $$

Factor each polynomial. $$ 125 x^{3}-216 $$

Factor each polynomial. $$ 125 y^{6}+z^{3} $$

Factor each trinomial. \(4(m-5)^{2}-4(m-5)-15\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

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.