Problem 92 - College Algebra Essentials

Question

Solve each equation in Exercises \(83-108\) by the method of your choice. $$ (2 x-5)(x+1)=2 $$

Step-by-Step Solution

Answer

The solutions for \( x \) in the equation \( (2 x - 5)(x + 1) = 2 \) are \( x = \frac{3 \pm \sqrt{33}}{4} \).

Step 1: Expand the expression

Apply the distributive (or FOIL) property to the expression on the left side of the equation. Multiply every term in the first parentheses by every term in the second parentheses. This results in: \( 2x^2 + 2x - 5x -5 = 2 \)

Step 2: Simplify the equation

Combine like terms on the left side of the equation resulting in the following: \( 2x^2 - 3x -5 = 2 \)

Step 3: Rearrange the equation

Rearrange the equation to equate to zero, done by subtracting 2 from both sides of the equation: \( 2x^2 - 3x - 7 = 0 \)

Step 4: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve the quadratic equation, where \( a = 2, b = -3, c = -7 \). Plugging these values into the quadratic formula will give the roots of the equation, which are the solutions for \( x \).

Key Concepts

Algebraic Expressions

When facing algebra homework, it's fundamental to understand algebraic expressions, which are combinations of variables, numbers, and operations. They form the backbone of algebra and enable us to describe relationships and changes in values mathematically. For instance, an expression like \((2x - 5)(x + 1)\)

is a product of two binomials, where variables (x) and constants (-5 and 1) are combined using addition and multiplication. To simplify, one can expand this expression, which involves distributing each term of the first polynomial to each term of the second polynomial, as done in the given exercise. Understanding how to correctly expand and combine like terms is crucial since it leads us to a form that can be used further for solving equations.

Quadratic Formula

The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is a powerful tool that provides a method to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). It comes from the process of completing the square in the derivation of the formula. The variables a, b, and c in the formula represent the coefficients of the equation where \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\), and \(c\) is the constant term.

This method is especially useful when factorization is difficult or not readily apparent. As seen in the exercise, once the quadratic equation is in standard form, you can identify a, b, and c, and then plug them into the formula to find the values of x which satisfy the equation. The \(\pm\) sign indicates that there will generally be two solutions for x, derived from the square root component of the formula.

Factoring Polynomials

Factoring is the process of breaking down a polynomial into its constituent factors such that when multiplied together, they give back the original polynomial. It's analogous to breaking a number down into its prime factors. For quadratic equations such as \(ax^2 + bx + c = 0\), one common method of factoring is to look for two binomials that when multiplied give the original quadratic. This is often possible when the quadratic is factorable over the integers.

The process involves finding two numbers that multiply to give ac (the product of the coefficients of \(x^2\) and the constant term) and add to give b (the coefficient of x). These numbers are used to split the middle term and factor by grouping. However, not all quadratics are easily factored, and that's when alternative methods such as the quadratic formula are invaluable.