Problem 82 - College Algebra

Question

Solve each equation in Exercises 73-98 by the method of your choice. \((2 x-5)(x+1)=2\)

Step-by-Step Solution

Answer

The solutions to the equation are \(x = \frac{3 + \sqrt{33}}{4}\) and \(x = \frac{3 - \sqrt{33}}{4}\).

Step 1: Expand the Left Side of the Equation

Expand \((2x-5)(x+1)\). Apply the distributive property to obtain \(2x^2-5x+2x-5\), which simplifies to \(2x^2-3x-5\). So, the equation becomes \(2x^2-3x-5=2\).

Step 2: Rearrange the Equation to Standard Quadratic Form

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). So, subtract 2 from both sides of the equation to obtain \(2x^2-3x-5-2=0\), which simplifies to \(2x^2-3x-3=0\).

Step 3: Apply the Quadratic Formula

The quadratic formula is \(-b ± \sqrt{b^2 - 4ac} / (2a)\). Substituting \(a = 2\), \(b = -3\), and \(c = -3\) into the formula, the roots of the equation are \[x = \frac{-(-3) ± \sqrt{(-3)^2 - 4*2*(-3)}}{2*2}\], which simplifies to \[x = \frac{3 ± \sqrt{9 +24}}{4} = \frac{3 ± \sqrt{33}}{4}\].

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). The general solution for the roots of such an equation is given by the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula provides the values of \(x\) for which the equation is satisfied. It's a universal method that works for all quadratic equations, as long as \(a\), \(b\), and \(c\) are real numbers, and \(a \eq 0\).

In our exercise, applying the quadratic formula to the equation \(2x^2-3x-3=0\) involved substituting \(a = 2\), \(b = -3\), and \(c = -3\). After computing the discriminant \(b^2 - 4ac\) and placing the values in the formula, we found the roots of the equation: \(\frac{3 \pm \sqrt{33}}{4}\). It's critical to remember both plus and minus signs lead to two possible solutions for \(x\), reflecting the parabolic nature of quadratic equations, which may intersect the x-axis at up to two points.

Distributive Property

The distributive property is a fundamental concept in algebra that allows us to multiply a single term by each term inside a parenthesis. The standard format for the distributive property is \(a(b + c) = ab + ac\).

To see the distributive property in action, let's take a look at the initial equation from our exercise: \(2x-5)(x+1)\). When we expand this expression, we multiply \(2x\) by both \(x\) and \(1\), and then \( -5\) by both \(x\) and \(1\). This leads us to \(2x^2 + 2x - 5x - 5\), which simplifies to \(2x^2 - 3x - 5\), as observed in Step 1 of the solution. Mastery of this property is crucial because it's used not only in solving quadratic equations but in simplifying expressions and solving various other types of equations throughout algebra.

Standard Quadratic Form

A quadratic equation is in standard form when it is expressed as \(ax^2 + bx + c = 0\). Understanding this form is important as it prepares the equation for various solution methods, including factoring, completing the square, or using the quadratic formula.

In our problem, we began with an equation that was not in standard form. By applying the distributive property, we expanded the equation to get \(2x^2-3x-5\). To put it in standard form, we then had to rearrange the equation to move all terms to one side, resulting in \(2x^2 - 3x - 3 = 0\) after subtracting \(2\) from both sides. Once in standard form, we were well-positioned to apply the quadratic formula and find the solutions for \(x\). This essential first step, converting the equation to standard form, sets the stage for all subsequent steps in solving quadratic problems.