Problem 47 - A Graphical Approach to College Algebra

Question

Solve each equation and inequality. (a) \(\frac{(2 x+1)(2 x)-\left(x^{2}+1\right)(2)}{(2 x+1)^{2}}=0\) (b) \(\frac{(2 x+1)(2 x)-\left(x^{2}+1\right)(2)}{(2 x+1)^{2}}<0\)

Step-by-Step Solution

Answer

(a) \(x = \frac{-1 \pm \sqrt{5}}{2}\); (b) \(x \in \left( \frac{-1 - \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2} \right)\).

Step 1: Simplify the Numerator

For both parts, begin by simplifying the numerator, \((2x+1)(2x) - (x^2+1)(2)\). Compute each part separately:\[(2x+1)(2x) = 4x^2 + 2x\]\[(x^2+1)(2) = 2x^2 + 2\]So the numerator becomes:\[4x^2 + 2x - (2x^2 + 2) = 4x^2 + 2x - 2x^2 - 2 = 2x^2 + 2x - 2\]

Step 2: Simplify the Expression

Substitute the simplified numerator back into the original expressions:(a) \(\frac{2x^2 + 2x - 2}{(2x+1)^2} = 0\)(b) \(\frac{2x^2 + 2x - 2}{(2x+1)^2} < 0\)\\Next, focus on these expressions as solution steps are different for each case.

Step 3: Solve the Equation for Part (a)

For the equation \(\frac{2x^2 + 2x - 2}{(2x+1)^2} = 0\), a fraction is zero when its numerator is zero. Solve the numerator:\[2x^2 + 2x - 2 = 0\]Factor it:\[2(x^2 + x - 1) = 0\]Factor \(x^2 + x - 1\):Using the quadratic formula for solving \(x^2 + x - 1 = 0\):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2}\]Thus, the solutions to part (a) are:\[x = \frac{-1 + \sqrt{5}}{2} \text{ and } x = \frac{-1 - \sqrt{5}}{2}\]

Step 4: Solve the Inequality for Part (b)

For \(\frac{2x^2 + 2x - 2}{(2x+1)^2} < 0\), analyze where the simplified numerator, \(2x^2 + 2x - 2 = 0\), is less than zero, considering that the denominator \((2x+1)^2\) is always positive.Find the critical points:\[x = \frac{-1 + \sqrt{5}}{2}, x = \frac{-1 - \sqrt{5}}{2}\]These critical points are where the numerator changes sign. Test intervals between and beyond these points:When \(x < \frac{-1 - \sqrt{5}}{2}\), test \(x = -3\), resulting in a positive value.When \(\frac{-1 - \sqrt{5}}{2} < x < \frac{-1 + \sqrt{5}}{2}\), test \(x = 0\), resulting in a negative value.When \(x > \frac{-1 + \sqrt{5}}{2}\), test \(x = 2\), resulting in a positive value.The inequality is satisfied over the interval:\[ \left( \frac{-1 - \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2} \right) \]

Key Concepts

Inequalities

Inequalities are expressions that show the relationship between two values that are not equal. They involve symbols such as < (less than), > (greater than), and ≤ or ≥ (less than or equal to, greater than or equal to respectively). Solving inequalities often involves finding the range of values for which the inequality holds true.
Understanding the nature of inequalities is crucial since they appear in multiple mathematical contexts. For example, in Exercise (b), the inequality was \( \frac{2x^2 + 2x - 2}{(2x+1)^2} < 0 \), which means we need to find the values of \(x\) that make the entire fraction less than zero.
To solve it, identify where the numerator \(2x^2 + 2x - 2\) is less than zero, ensuring the denominator remains positive since it is a squared term \((2x+1)^2\). Inequalities require testing intervals around the critical points derived from the equation of the numerator. By doing this, you can determine where the expression changes signs and find intervals of \(x\) that satisfy the inequality.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which take the general form \(ax^2 + bx + c = 0\). When the equation does not factorize easily, the quadratic formula comes to the rescue:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula gives both potential solutions to the quadratic equation by considering both the '+' and '−' operations in \(\pm\).
In the original exercise, the quadratic formula was used to solve \(x^2 + x - 1 = 0\), found in the numerator after it was simplified in Step 3. Plugging in \(a = 1\), \(b = 1\), and \(c = -1\) into the formula gives us:\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-1)}}{2 \times 1} \] Simplifying the expression provides exact solutions for \(x\), guiding the resolution of the equation. Knowing how to apply the quadratic formula efficiently is integral to solving complex algebraic problems.

Factoring

Factoring is a technique used to simplify algebraic expressions or solve equations, usually by expressing a polynomial as a product of its factors. It is particularly useful for solving quadratic equations when the polynomial can be decomposed into simpler binomials.
In Step 3, after rewriting \(2x^2 + 2x - 2 = 0\) as \(2(x^2 + x - 1) = 0\), we aim to factor \(x^2 + x - 1\). Here, the quadratic does not factor into neat numbers, which is why the quadratic formula was utilized.
Despite not being factorable in this instance, attempting to factor a quadratic equation is often the first step in simplifying and solving, as it can lead to quicker solutions when the roots or factors are integers or simple fractions.

Critical Points

Critical points are essential in determining where and how functions change behavior, particularly with inequalities. They are values of \(x\) where the sign of a polynomial, such as the numerator of a fraction, changes. They are found by setting the related polynomial equal to zero and solving for \(x\).
For Exercise (b), we located the critical points \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\). These points indicate where the expression changes from positive to negative or vice versa.
Testing intervals determined by these critical points is essential to solve inequalities because it identifies the ranges where the inequality holds. In this specific exercise, checking the signs of test values in each interval determined the part of the number line that satisfied the inequality condition.