Question

Find the zero(s) of the function. (See Example 4.) \(h(x)=x^2+19 x+84\)

Short answer

The zeros of the function \(h(x) = x^2 + 19x + 84\) are \(x = -7\) and \(x = -12\).

Step-by-step solution

Identify the coefficients

In the given function, \(h(x) = x^2 + 19x + 84\), the coefficients are as follows: \(a = 1\) (coefficient of \(x^2\)), \(b = 19\) (coefficient of \(x\)), and \(c = 84\) (constant term).

Substitute the coefficients into the quadratic formula

Substitute the coefficients \(a\), \(b\), and \(c\) into the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This gives \(x = \frac{-19 \pm \sqrt{19^2 - 4*1*84}}{2*1}\).

Simplify under the square root

First, simplify the operation under the square root: \(19^2 - 4*1*84 = 361 - 336 = 25\).

Substitute into formula

Substitute this result back into the formula: \(x = \frac{-19 \pm \sqrt{25}}{2}\)

Simplify further

Square root of 25 is 5, now the equation is \(x = \frac{-19 \pm 5}{2}\). We have two possible solutions here: \(x = \frac{-19 + 5}{2}\) and \(x = \frac{-19 - 5}{2}\)

Solve for x

Solve these two equations to find the values of x. The solutions are \(x = -7\) and \(x = -12\)

Key concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). This formula provides a reliable way to find the values of \(x\) that satisfy the equation, commonly known as the roots. The quadratic formula is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Let's break it down:

• \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation, representing the coefficients of \(x^2\), \(x\), and the constant term respectively.
• The symbol \(\pm\) indicates that the formula considers both addition and subtraction, leading to two potential solutions.
• The part under the square root, \(b^2 - 4ac\), is called the discriminant. It determines the nature of the roots.

By substituting the values of \(a\), \(b\), and \(c\) from the given problem into the formula, you can solve the equation step by step.

Zeros of a Function

Zeros of a function refer to the values of \(x\) at which the function \(f(x)\) equals zero. In other words, they are the points where the graph of the function crosses or touches the x-axis.

• For a quadratic function like \(h(x) = x^2 + 19x + 84\), finding the zeros is equivalent to solving the equation \(h(x) = 0\).

To find these zeros, we set the equation \(x^2 + 19x + 84 = 0\) and apply the quadratic formula:

• We found that \(x = -12\) and \(x = -7\), meaning these are the x-values where the function's output is zero.

These solutions not only help in solving equations but also in understanding the behavior of quadratic graphs.

Solving Equations

Solving equations involving quadratic expressions is a frequent requirement in algebra. It allows us to find unknown values that satisfy certain conditions described by an equation.

• Start by identifying the coefficients \(a\), \(b\), and \(c\) in the equation.
• Use the quadratic formula to perform calculations efficiently, ensuring all operations are simplified correctly.

Through our example, substituting into the formula gave us:

• \(x = \frac{-19 \pm \sqrt{25}}{2}\)

Where \(\sqrt{25}\) equals 5, leading us to the final solutions. This process enhances skills in algebraic manipulation and solution verification.

Algebra 2 Concepts

Algebra 2 introduces more complex topics and advanced techniques for solving equations. Concepts from Algebra 2 build on the foundation established in Algebra 1, broadening problem-solving capabilities. Key concepts include:

• Polynomials and their properties, such as degree and zeros.
• Quadratic equations and different methods to solve them, including factoring, completing the square, and using the quadratic formula.
• Understanding the discriminant's role in determining the nature of roots.

In our problem, the quadratic formula was the chosen method, showcasing its effectiveness even when equations have no simple factorization. Mastering these concepts helps in tackling real-world applications and prepares students for further mathematical studies.