Problem 174 - 1,001 GED Practice Questions For Dummies

Question

Which of the following lists the solutions to the equation \(x^2-x-20=0\) ? (A) 2 and \(-10\) (B) \(-2\) and 10 (C) 4 and \(-5\) (D) \(-4\) and 5

Step-by-Step Solution

Answer

The short answer is: The solutions to the equation \(x^2 - x - 20 = 0\) are \(x = -4\) and \(x = 5\), which corresponds to option (D).

Step 1: Identify the quadratic equation

Given quadratic equation is \(x^2 - x - 20 = 0\).

Step 2: Factor the quadratic equation

We need two numbers whose product is equal to the constant term (-20) and whose sum is equal to the coefficient of x (-1). We can choose the numbers -5 and 4 because their product is -20 and their sum is -1: \((x - 5)(x + 4) = 0\).

Step 3: Solve for x

To find the solutions, we will solve the factorized equation for x by setting each factor equal to zero, as follows: 1. \(x - 5 = 0\) 2. \(x + 4 = 0\)

Step 4: Identify the solution

1. \(x - 5 = 0 \Rightarrow x = 5\) 2. \(x + 4 = 0 \Rightarrow x = -4\) So the solutions are 5 and -4, which corresponds to option (D). Therefore, the correct answer is: **(D) -4 and 5**

Key Concepts

Factoring Quadratics

Factoring a quadratic equation involves expressing it as a product of two binomials. This process is useful because it transforms the quadratic equation into a simpler form that can be easily solved. Let's break down the process of factoring quadratics using the equation from the exercise:

• The given equation is \(x^2 - x - 20 = 0\).
• We need to find two numbers that multiply to give the constant term, which is \(-20\), and add up to the coefficient of \(x\), which is \(-1\).
• By examining potential pairs, we find that the numbers \(-5\) and \(4\) fit the criteria, as \(-5 \times 4 = -20\) and \(-5 + 4 = -1\).

These numbers allow us to factor the quadratic equation into \((x - 5)(x + 4) = 0\). By transforming the original quadratic equation into this factored form, we make it ready for solving.

Solutions of Quadratic Equations

Once we have factored a quadratic equation, finding the solutions becomes straightforward. The solutions of a quadratic equation are the values of \(x\) that satisfy the equation. Let's see how we achieve this:

• From the factored equation \((x - 5)(x + 4) = 0\), we set each binomial to zero: \(x - 5 = 0\) and \(x + 4 = 0\).
• By solving these equations, we obtain possible solutions for \(x\).
• For \(x - 5 = 0\), adding \(5\) to both sides gives \(x = 5\).
• For \(x + 4 = 0\), subtracting \(4\) from both sides gives \(x = -4\).

Thus, the solutions are \(x = 5\) and \(x = -4\). These solutions indicate the points where the quadratic function intersects the x-axis when graphically represented.

Solving Equations

Solving equations is a fundamental skill in mathematics. It involves determining the value of the variable that makes the equation true. Let's recap how this is applied to our quadratic equation:

• We start with the original quadratic equation \(x^2 - x - 20 = 0\).
• By factoring it into \((x - 5)(x + 4) = 0\), we simplify the equation. This makes each factor equal to zero a simple linear equation.
• Solving the linear equations \(x - 5 = 0\) and \(x + 4 = 0\) separately, we find the solutions that satisfy the original equation.
• These solutions are the values for \(x\) that make the equation true, specifically \(x = 5\) and \(x = -4\).

Understanding how to solve these equations allows us to find where functions cross the x-axis, an important concept in algebra and calculus.