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Chapter 4: Problem 1

For the function \(f(x)=x^{2}+2 x-15,\) solve each of the following. $$f(x)=0$$

Short Answer

Expert verified
The solutions are \(x = -5\) and \(x = 3\).

Step by step solution

01

Set the function equal to zero

We start by setting the given function equal to zero.Given: \[ f(x) = x^2 + 2x - 15 \]To solve for the values of \(x\) where \(f(x) = 0\), set the function equal to zero:\[ x^2 + 2x - 15 = 0 \]
02

Factor the quadratic equation

Next, factor the quadratic equation. We look for two numbers that multiply to -15 and add to 2. These numbers are 5 and -3.Thus, the quadratic equation can be factored as:\[ (x + 5)(x - 3) = 0 \]
03

Solve for x

Now that we have factored the equation, set each factor equal to zero and solve for \(x\).\[ x + 5 = 0 \]Subtract 5 from both sides:\[ x = -5 \]\[ x - 3 = 0 \]Add 3 to both sides:\[ x = 3 \]
04

Verify the solutions

To verify the solutions, substitute \(x = -5\) and \(x = 3\) back into the original function and ensure \(f(x) = 0\).For \(x = -5\):\[ f(-5) = (-5)^2 + 2(-5) - 15 = 25 - 10 - 15 = 0 \]For \(x = 3\):\[ f(3) = (3)^2 + 2(3) - 15 = 9 + 6 - 15 = 0 \]Both solutions satisfy the equation.

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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 fundamental skill in algebra. It involves expressing a quadratic equation in the form of \(ax^2 + bx + c = 0\), as a product of two binomial expressions. This can make solving the equation simpler. In this instance, we started with \( x^2 + 2x - 15 = 0 \). To factor this, we look for two numbers that multiply to -15 (the constant term) and add up to 2 (the coefficient of the linear term). These numbers are 5 and -3, giving us the factored form \( (x + 5)(x - 3) = 0 \). Factoring is essentially looking for roots that satisfy the quadratic equation.
Quadratic Equations
Quadratic equations are second-degree polynomial equations of the form \( ax^2 + bx + c = 0 \). They frequently appear in algebra and describe parabolic shapes when graphed. The solutions to quadratic equations, known as the roots, can be found in several ways:
  • Factoring
  • Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
  • Completing the square
In our problem, we solved \( x^2 + 2x - 15 = 0 \) by factoring. Understanding this process requires recognizing that the solutions to \( x^2 + 2x - 15 = 0 \) occur where the quadratic expression equals zero: these are the points where the parabola crosses the x-axis.
Verification of Solutions
Verifying solutions is a crucial step in solving equations. It ensures that the solutions obtained are correct. For quadratic equations, this involves substituting the solutions back into the original equation. In our example, we found \( x = -5 \) and \( x = 3 \). To verify, substitute these values back into the original equation \( x^2 + 2x - 15 \):
  • For \( x = -5 \): \( (-5)^2 + 2(-5) - 15 = 25 - 10 - 15 = 0 \)
  • For \( x = 3 \): \( (3)^2 + 2(3) - 15 = 9 + 6 - 15 = 0 \)
Since both computations result in zero, this confirms that \( x = -5 \) and \( x = 3 \) are indeed the correct solutions. This step provides confidence in the methods used and ensures accuracy.

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