Problem 54 - Beginning and Intermediate Algebra

Question

Solve each equation. $$-84=s(s+19)$$

Step-by-Step Solution

Answer

The solutions of the equation \(-84 = s(s+19)\) are \(s=-7\) and \(s=-12\).

Step 1: Expand the given equation

In this step we will multiply s with each term in the parentheses \((s+19)\), so the equation becomes: $$-84 = s^2 + 19s$$

Step 2: Rewrite the equation in standard form

The standard form for a quadratic equation is $$ax^2 + bx + c = 0$$. Therefore, we need to bring all the terms to one side of the equation: $$s^2 + 19s + 84 = 0$$

Step 3: Use the quadratic formula to find the values of s

The quadratic formula is given by: $$s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our equation, we have $$a = 1$$, $$b = 19$$, and $$c = 84$$. #Phase 2#

Step 4: Plug the values of a, b, and c into the quadratic formula

Now, we will substitute the values of a, b, and c into the quadratic formula: $$s = \frac{-19 \pm \sqrt{19^2 - 4(1)(84)}}{2(1)}$$

Step 5: Simplify the equation

First, we can simplify the terms under the square root: $$s = \frac{-19 \pm \sqrt{361 - 336}}{2}$$ $$s = \frac{-19 \pm \sqrt{25}}{2}$$ Now, we will simplify the square root and solve for s: $$s_1 = \frac{-19 + 5}{2} = -7$$ $$s_2 = \frac{-19 - 5}{2} = -12$$ Therefore, the solutions of the equation $$-84 = s(s+19)$$ are \(s=-7\) and \(s=-12\).

Key Concepts

Expanding Algebraic Expressions

When solving quadratic equations, the first step often involves expanding algebraic expressions. This means multiplying each term in one expression by each term in another.
In our exercise, we are given an equation in the form:

• \(-84 = s(s + 19)\)

To expand this expression, multiply the variable \(s\) by each term inside the parentheses.
This results in:

• \(s \times s = s^2\)
• \(s \times 19 = 19s\)

Thus, the expanded form of the equation is \(-84 = s^2 + 19s\).
Once we have expanded the expression, we proceed to work with an equation in its quadratic form.

Solving Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). In our situation, after expanding, the quadratic equation is \(s^2 + 19s + 84 = 0\). To solve this, it is necessary to reformulate the equation into the standard quadratic form where all terms are on one side with zero on the other. Quadratic equations can be solved using different methods, such as factoring, completing the square, or using the quadratic formula.
In this exercise, we use the quadratic formula due to its convenience and because factoring might not be straightforward. The main objective is to find values of \(s\) that satisfy the equation.

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as:\[s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our equation \(s^2 + 19s + 84 = 0\), the coefficients are \(a = 1\), \(b = 19\), and \(c = 84\).
Replacing these in the quadratic formula gives:

• \(s = \frac{-19 \pm \sqrt{19^2 - 4 \times 1 \times 84}}{2 \times 1}\)

We simplify under the square root first. This gives:

• \(\sqrt{361 - 336} = \sqrt{25}\)
• \(\sqrt{25} = 5\)

So, substituting back, we have two solutions for \(s\):

• \(s_1 = \frac{-19 + 5}{2} = -7\)
• \(s_2 = \frac{-19 - 5}{2} = -12\)

These solutions, \(s = -7\) and \(s = -12\), satisfy the original equation.