Problem 35 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ x^{2}-3 x+6=0 $$

Step-by-Step Solution

Answer

\( x = \frac{3 + i\sqrt{15}}{2} \) and \( x = \frac{3 - i\sqrt{15}}{2} \).

Step 1: Identify coefficients

The quadratic equation is given in the standard form \( ax^2 + bx + c = 0 \). In this case, \( a = 1 \), \( b = -3 \), and \( c = 6 \).

Step 2: Write the quadratic formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 3: Substitute coefficients into the formula

Insert the values of \( a \), \( b \), and \( c \) into the quadratic formula: \( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4\cdot 1 \cdot 6}}{2\cdot 1} \). Simplifying the terms, we get: \( x = \frac{3 \pm \sqrt{9 - 24}}{2} \).

Step 4: Simplify under the square root

Calculate the term inside the square root: \( b^2 - 4ac = 9 - 24 = -15 \). Now the equation becomes: \( x = \frac{3 \pm \sqrt{-15}}{2} \).

Step 5: Express the square root of negative number

Rewrite the square root of the negative number in terms of imaginary units: \( \sqrt{-15} = \sqrt{15}i \). Now the equation is: \( x = \frac{3 \pm i\sqrt{15}}{2} \).

Step 6: Write the final solutions

Write out the two complex solutions as: \( x = \frac{3 + i\sqrt{15}}{2} \) and \( x = \frac{3 - i\sqrt{15}}{2} \).

Key Concepts

Complex Numbers

Complex numbers are an extension of the real number system. They involve a combination of a real part and an imaginary part. The general form is written as \(a + bi\). Here, \(a\) is the real part and \(bi\) is the imaginary part. These numbers arise when we solve equations that do not have real solutions. For instance, \(x^2 + 1 = 0\) leads to \(x = \pm i\) because there's no real number whose square is \(-1\).
Complex numbers are important because they help in solving quadratic equations that result in negative numbers under the square root.

Solving Quadratic Equations

To solve a quadratic equation like \(ax^2 + bx + c = 0\), we often use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula helps find the roots of any quadratic equation by inserting the coefficients \(a\), \(b\), and \(c\). Let's break down the steps using the given example.

• Identify coefficients: For \(x^2 - 3x + 6 = 0\), we have \(a = 1\), \(b = -3\), and \(c = 6\).
• Insert these coefficients into the formula: \(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}\).
• Simplify the terms: This leads to \(x = \frac{3 \pm \sqrt{9 - 24}}{2}\).
• Simplify under the square root: We get \(9 - 24 = -15\), so the number under the square root is \(-15\).

Imaginary Units

Imaginary units are used to handle the square roots of negative numbers. The imaginary unit, denoted by \(i\), is defined as \(i = \sqrt{-1}\). Using \(i\), we can express the square root of any negative number. If you come across \(\sqrt{-a}\) where \(a\) is positive, it can be written as \(\sqrt{a}i\).
In our example, we encounter \(\sqrt{-15}\). Using the imaginary unit, we express it as \(\sqrt{15}i\). This converts the quadratic equation to \(x = \frac{3 \pm \sqrt{15}i}{2}\). This way, we can represent results that are not part of the real number system.

Quadratic Equation Coefficients

In a general quadratic equation \(ax^2 + bx + c = 0\), \(a\), \(b\), and \(c\) are known as the coefficients.

• \(a\) is the coefficient of \(x^2\) (the quadratic term).
• \(b\) is the coefficient of \(x\) (the linear term).
• \(c\) is the constant term.

For the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), these coefficients help determine the solutions.
Let’s take the example provided: \(x^2 - 3x + 6 = 0\). Here, \(a = 1\), \(b = -3\), and \(c = 6\). By substituting them into the quadratic formula, we solve for \(x\). The coefficients tell us how the parabola opens and shifts on a graph. They are essential in finding roots and understanding the shape of the related quadratic function.