Problem 64 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation using any method. Find only real solutions. $$(x+1)(x-2)=2$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = {1 + \sqrt{17} \over 2}\) and \(x = {1 - \sqrt{17} \over 2}\).

Step 1: Expand the equation

Multiply the two factors on the left side to get \(x^2 - 2x + x - 2 = 2\). This simplifies to \(x^2 - x - 2 = 2\).

Step 2: Rearrange into standard form

Subtract 2 from both sides to make the equation equal to zero, which gives us \(x^2 - x - 4 = 0\).

Step 3: Use quadratic formula

The quadratic formula is \(x = {-b ± \sqrt{b^2 - 4ac} \over 2a}\). Substituting \(a=1\), \(b=-1\), and \(c=-4\) into the formula, we get: \(x = {1 ± \sqrt{1+16} \over 2} = {1 ± \sqrt{17} \over 2}\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are mathematical expressions in the form of \(ax^2 + bx + c = 0\). This particular formula provides a systematic way to find the roots of any quadratic equation.

The general form of the quadratic formula is:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
When solving an equation using the quadratic formula, you first identify the coefficients \(a\), \(b\), and \(c\) from the equation. Next, you substitute these values into the formula. The term under the square root, \(b^2 - 4ac\), is known as the discriminant. The discriminant determines the nature of the roots. When \(b^2 - 4ac > 0\), there are two distinct real roots. If \(b^2 - 4ac = 0\), the equation has one real root (a repeated root). If \(b^2 - 4ac < 0\), the equation has no real solutions but two complex solutions.

Applying the quadratic formula to the sample equation \(x^2 - x - 4 = 0\), we use \(a=1\), \(b=-1\), and \(c=-4\), leading to the real roots \(\frac{{1 \pm \sqrt{17}}}{{2}}\).

Factoring Quadratics

Factoring quadratics is another method to solve quadratic equations. The process involves rewriting the quadratic in the form of \( (x - p)(x - q) = 0\), where \(p\) and \(q\) are the roots of the equation. Once factored, you can easily solve for \(x\) by setting each factor equal to zero and solving for \(x\).

However, not all quadratics are factorable with rational numbers. In cases where the solutions are not rational, we must use other methods such as completing the square or applying the quadratic formula.

Factoring can be a quicker method when the roots are rational numbers, and once the quadratic is adequately factored, it's simple to find the zeros of the function. It is a useful technique to understand as it helps build a deeper understanding of the nature of quadratic equations.

Real Solutions of Quadratics

Real solutions of quadratics refer to the \(x\)-values that satisfy the quadratic equation \(ax^2 + bx + c = 0\) when plotted on a graph. These solutions are also known as \(x\)-intercepts or zeros of the function.

The number of real solutions can be determined by examining the discriminant (\(b^2 - 4ac\)). If the discriminant value is positive, there are two different real solutions. When it's exactly zero, the quadratic has exactly one real solution, which is also called a double root. If the discriminant is negative, there are no real solutions; the solutions are complex numbers.

In the exercise given, using the quadratic formula revealed that the discriminant was positive (since \(1^2 - 4(1)(-4) = 17\)), indicating two distinct real solutions. For quadratics that we can factor, if we can find two numbers that multiply to \(ac\) and add to \(b\), we can find the real solutions easily by setting each factor equal to zero and solving.