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Chapter 1: Problem 21

Solve each equation by factoring. $$ x(x+3)-10=0 $$

Short Answer

Expert verified
The solutions are \( x = -5 \) and \( x = 2 \).

Step by step solution

01

- Expand the equation

Start by expanding the equation if necessary. In this case, distribute the x: \[ x(x + 3) - 10 = 0 \] This simplifies to \[ x^2 + 3x - 10 = 0 \]
02

- Identify the factors

You need to find two numbers that multiply to -10 and add up to 3. These numbers are 5 and -2, because \[ 5 \times (-2) = -10 \] and \[ 5 + (-2) = 3 \]
03

- Write the equation in factored form

Using the numbers found, write the equation in its factored form: \[ (x + 5)(x - 2) = 0 \]
04

- Solve for x

Set each factor equal to 0 and solve for x: \[ x + 5 = 0 \] leads to \[ x = -5 \] and \[ x - 2 = 0 \] leads to \[ x = 2 \]

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Key Concepts

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

Factoring Trinomials
Factoring trinomials is a crucial skill in algebra, especially when you're working to solve quadratic equations. Let's break down the original trinomial we had in the problem:

The equation was simplified to:
\[ x^2 + 3x - 10 = 0 \]

Now, to factor this trinomial, you鈥檙e looking for two numbers that multiply to the constant term (-10) and add up to the coefficient of the middle term (3). Here鈥檚 how you tackle it:
  • Write down the factors of -10: {-10, 1}, {5, -2}, {-5, 2}, {10, -1}
  • Identify the pair that adds up to 3. In this case, it鈥檚 (5, -2).
Using these numbers, you can write the trinomial in factored form as:
\[ (x + 5)(x - 2) = 0 \]

This is how you convert the quadratic trinomial into a product of two binomials.
Solving Quadratic Equations
Once you have factored your quadratic equation, solving it becomes straightforward. Recall our factored equation:
\[ (x + 5)(x - 2) = 0 \]

The Zero Product Property states that if the product of two factors is zero, at least one of the factors must be zero. So, you can set each factor to zero and solve for x:
  • \[ x + 5 = 0 \] leads to \[ x = -5 \]
  • \[ x - 2 = 0 \] leads to \[ x = 2 \]
So, the solutions to the original equation are:
\[ x = -5 \] and \[ x = 2 \]

This means these values of x make the equation true when substituted back.
Algebraic Expressions
Understanding algebraic expressions is fundamental before diving into factoring and solving equations. An algebraic expression involves variables, coefficients, and constants combined using arithmetic operations. In the exercise at hand, you began with:
\[ x(x + 3) - 10 = 0 \]

Expanding it to form a standard quadratic expression, it becomes:
\[ x^2 + 3x - 10 = 0 \]

Here:
  • \(x^2\) is the quadratic term.
  • \(3x\) is the linear term.
  • \(-10\) is the constant term.
By manipulating these terms through factoring, you simplify the equation into something that鈥檚 easier to solve. Mastery over handling algebraic expressions is the stepping stone to solving more complex equations.

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Most popular questions from this chapter

Find the real solutions, if any, of each equation. Use any method. $$ x^{2}+\sqrt{2} x=\frac{1}{2} $$

Challenge Problem If \(\frac{2}{3} \leq \frac{5-x}{3} \leq 3,\) find the largest possible value of \(2 x^{2}-3\)

In going from Chicago to Atlanta, a car averages 45 miles per hour, and in going from Atlanta to Miami, it averages 55 miles per hour. If Atlanta is halfway between Chicago and Miami, what is the average speed from Chicago to Miami? Discuss an intuitive solution. Write a paragraph defending your intuitive solution. Then solve the problem algebraically. Is your intuitive solution the same as the algebraic one? If not, find the flaw.

Describe three ways that you might solve a quadratic equation. State your preferred method; explain why you chose it.

Do you prefer to use inequality notation or interval notation to express the solution to an inequality? Give your reasons. Are there particular circumstances when you prefer one to the other? Cite examples.
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