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Chapter 0: Problem 58

$$\text { solve each quadratic equation by factoring.}$$ $$x^{2}=-11 x-10$$

Short Answer

Expert verified
The solutions to the quadratic equation \(x^{2} + 11x + 10 = 0\) are \(x = -10\) and \(x = -1\).

Step by step solution

01

Set the equation to zero

To start factoring, it's necessary to have the equation set to zero: \(x^{2} + 11x + 10 = 0\)
02

Factoring the quadratic equation

Next, factor the quadratic equation. Look for two numbers that multiply to 10 (the constant term) and add up to 11 (the coefficient of x). Those numbers are 10 and 1. This gives: \((x + 10)(x + 1) = 0\)
03

Apply the Zero-Product Property

The Zero-Product Property states that if a product of factors equals zero, then at least one of the factors must be zero. Set each factor equal to zero and solve for x to find the solutions: \(x + 10 = 0\) and \(x + 1 = 0\). Solving these gives: \(x = -10\) and \(x = -1\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Quadratics
Factoring quadratics is a method used to solve quadratic equations by expressing the equation in a product form. The general form of a quadratic equation is \(ax^2 + bx + c = 0\). To factor it, we need to find two binomials that multiply to form the quadratic. Here's how you can do it:
  • First, ensure the equation is set to zero. This means re-arranging terms, if necessary, to have the quadratic expression equal to zero.

  • Next, identify the constant term \(c\) and the coefficient of \(x\), which is \(b\).

  • Find two numbers that multiply to \(c\) (the constant term) and add up to \(b\) (the coefficient of \(x\)). These two numbers will help break up the middle term in order to factor the quadratic.

In the example from the exercise, we had to solve \(x^2 + 11x + 10 = 0\). The numbers 10 and 1 were determined because they multiply to 10 and add up to 11, allowing us to factor it as \((x + 10)(x + 1) = 0\). Understanding how to factor quadratics is crucial because it allows us to find the solutions of the equation easily.
Zero-Product Property
The Zero-Product Property is a fundamental concept in algebra, particularly when working with quadratic equations. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. This is essential for solving equations that have been factored.
  • Once you've factored the quadratic equation, set each factor equal to zero.

  • Solving each of these simple equations will yield the roots or solutions to the original quadratic equation.

For the exercise's example, after factoring \(x^2 + 11x + 10 = 0\) into \((x + 10)(x + 1) = 0\), you apply the Zero-Product Property:
  • Set \(x + 10 = 0\), which gives \(x = -10\).

  • Set \(x + 1 = 0\), which gives \(x = -1\).

This property is instrumental because it transforms a potentially complex problem into a series of simpler linear equations.
Polynomial Equations
Polynomial equations are equations that involve terms with variables raised to whole-number exponents. A quadratic equation is a type of polynomial equation with the highest exponent being 2. Solving polynomial equations requires a good understanding of various algebraic techniques, including factoring and the Zero-Product Property.
  • Quadratic equations, like the exercise example \(x^2 + 11x + 10 = 0\), are solved by detecting the polynomial structure and then using methods like factoring.

  • Factoring is just one way to solve polynomial equations. Other methods include using the quadratic formula or completing the square.

Understanding polynomial equations is critical, as they can model many real-world phenomena. Recognizing the different methods to solve them allows for flexibility and accuracy in finding solutions. By mastering these concepts, such as factoring and the zero-product property, solving polynomial equations becomes a systematic process.

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Most popular questions from this chapter

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. You are choosing between two texting plans. Plan A has a monthly fee of \(\$ 15\) with a charge of \(\$ 0.08\) per text. Plan \(\mathbf{B}\) has a monthly fee of \(\$ 3\) with a charge of \(\$ 0.12\) per text. How many text messages in a month make plan A the better deal?

What is the discriminant and what information does it provide about a quadratic equation?

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. A truck can be rented from Basic Rental for \(\$ 50\) per day plus \(\$ 0.20\) per mile. Continental charges \(\$ 20\) per day plus \(\$ 0.50\) per mile to rent the same truck. How many miles must be driven in a day to make the rental cost for Basic Rental a better deal than Continental's?

Perform the indicated operations. Simplify the result, if possible. $$\frac{1}{x^{2}-2 x-8} \div\left(\frac{1}{x-4}-\frac{1}{x+2}\right)$$

Explain how to simplify a rational expression.
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