/*! 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 34 Write a quadratic equation havin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 34

Write a quadratic equation having the given numbers as solutions. $$ -2,-5 $$

Short Answer

Expert verified
The quadratic equation with solutions -2 and -5 is \[ x^2 + 7x + 10 = 0 \].

Step by step solution

01

Understand the given solutions

A quadratic equation can be written as If -2 and -5 are the solutions, they are the values of x that satisfy the equation.
02

Use the form of the quadratic equation

The standard form of a quadratic equation with solutions x = a and x = b is given by: The solutions given are x = -2 and x = -5, so we substitute a = -2 and b = -5.
03

Write the factors

The factors of the quadratic equation are written using the solutions: \[ (x + 2)(x + 5) = 0 \] Since -2 and -5 are the solutions, they make each of the factors zero.
04

Expand the factors to form the quadratic equation

Expand the factors: \[ (x + 2)(x + 5) = x^2 + 5x + 2x + 10 \] Combine like terms: \[ x^2 + 7x + 10 \] The quadratic equation in standard form is: \[ x^2 + 7x + 10 = 0 \]

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 solutions
Quadratic solutions are the values that satisfy a quadratic equation. If you are given the solutions of a quadratic equation, it means you have the roots of the equation. For example, if we have solutions \(-2\) and \(-5\), it means that these are the values of \(x\) that make the equation equal to zero.
In other words, when these values are plugged into the quadratic equation, the result is zero.
To write a quadratic equation based on given solutions, we must identify these solutions and use them to form factors.
factorization
Factorization is the process of breaking down an expression into a product of simpler expressions or factors. For quadratic equations, this involves finding the roots (solutions) and expressing the equation in its factorized form as \[ (x - a)(x - b) \].
Given the solutions \(-2\) and \(-5\), we write the factors of our quadratic equation as \[ (x + 2)(x + 5) = 0 \].
This method allows us to work backward from the solutions to the original quadratic equation by multiplying these factors together.
standard form of quadratic equation
The standard form of a quadratic equation is \[ ax^2 + bx + c = 0 \]. Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable.
In our problem, the given solutions are \(-2\) and \(-5\), and we need to convert these solutions into the standard form.
By writing it in the factorized form \[ (x + 2)(x + 5) = 0 \], we can expand it to find the equation in its standard form.
expanding factors
Expanding factors involves multiplying the factors to get the quadratic equation in its expanded form. From the factorized form of our solutions, \[ (x + 2)(x + 5) \], we carry out the multiplication:
First, distribute each term in the first binomial to every term in the second binomial:
- \( x \times x = x^2 \)
- \( x \times 5 = 5x \)
- \( 2 \times x = 2x \)
- \( 2 \times 5 = 10 \)
Combine the terms to get the expanded form:
\[ x^2 + 5x + 2x + 10 = x^2 + 7x + 10 \]
Thus, the standard form of the quadratic equation with solutions \(-2\) and \(-5\) is \[ x^2 + 7x + 10 = 0 \].

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. $$ 25 y^{2}+16=0 $$

Write an equation for a function having a graph with the same shape as the graph of \(f(x)=\frac{3}{5} x^{2},\) but with the given point as the vertex. $$ (-2,-5) $$

Let \(f(x)=x^{2} .\) Find \(x\) such that \(f(x)=11\).

Find the \(x\) -intercepts of the function given by \(f(x)=4 x^{2}+2 x-3\).

For each equation under the given condition, (a) find \(k\) and (b) find the other solution. Suppose that \(f(x)=a x^{2}+b x+c,\) with \(f(-3)=0, f\left(\frac{1}{2}\right)=0,\) and \(f(0)=-12 .\) Find \(a\) \(b,\) and \(c\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.