Problem 62 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation and express them in the form \(a+b i\) $$x^{2}-6 x+10=0$$

Step-by-Step Solution

Answer

The solutions are \(3 + i\) and \(3 - i\).

Step 1: Write Down the Quadratic Formula

The quadratic equation in standard form is given by \(ax^2 + bx + c = 0\). The solutions can be found using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] Here, \( a = 1\), \( b = -6\), and \( c = 10\).

Step 2: Calculate the Discriminant

The discriminant is given by the expression \(b^2 - 4ac\). Substitute the values \(b = -6\), \(a = 1\), and \(c = 10\): \[ (-6)^2 - 4 \times 1 \times 10 = 36 - 40 = -4.\] Since the discriminant is negative, the solutions will be complex numbers.

Step 3: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{-4}}{2 \times 1}.\] This simplifies to \[ x = \frac{6 \pm \sqrt{-4}}{2}.\]

Step 4: Simplify the Square Root of the Negative Discriminant

The expression \(\sqrt{-4}\) can be rewritten using imaginary numbers: \(\sqrt{-4} = \sqrt{4} \cdot i = 2i\). Thus, substitute back into the equation: \[ x = \frac{6 \pm 2i}{2}.\]

Step 5: Simplify the Expression

Separate into real and imaginary parts:\[ x = \frac{6}{2} \pm \frac{2i}{2}.\] This simplifies to \[ x = 3 \pm i.\] Thus the solutions in the form \(a + bi\) are \(3 + i\) and \(3 - i\).

Key Concepts

Complex Numbers

Complex numbers are numbers that have two parts: a real part and an imaginary part. They are generally expressed in the form \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part. The imaginary unit \(i\) is used to represent the square root of \(-1\), which means \(i^2 = -1\).

Complex numbers extend the concept of the regular number line into a plane, known as the complex plane. In this plane, the horizontal axis represents the real part and the vertical axis represents the imaginary part. Complex numbers are useful in various fields, including engineering, physics, and applied mathematics.

When solving quadratic equations, if the discriminant is negative, the solutions will be complex, indicating they cannot be represented on the real number line alone but require this two-dimensional space.

Quadratic Formula

The quadratic formula is a critical tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\).

The formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Using this formula, you can find the roots of the quadratic equation by substituting the values of \(a\), \(b\), and \(c\).

This formula works for all types of roots: real and distinct, real and repeated, and complex (when the discriminant is negative). We use the plus-minus symbol \(\pm\) because quadratic equations can have two solutions.

• The first solution is found by using the plus sign, \(-b + \sqrt{b^2 - 4ac}\), and
• the second is found using the minus sign, \(-b - \sqrt{b^2 - 4ac}\).

Discriminant

The discriminant is a specific part of the quadratic formula given by the expression \(b^2 - 4ac\).

It provides crucial information about the nature of the roots of a quadratic equation.

The value of the discriminant can tell us whether the roots are real or complex:

• If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root, known as a repeated or double root.
• If \(b^2 - 4ac < 0\), the roots are complex and occur as a conjugate pair.

For example, in our exercise, the discriminant is \(-4\), indicating the presence of complex roots, specifically imaginary numbers, as part of the solution.

Imaginary Numbers

Imaginary numbers arise from the square roots of negative numbers. The basic imaginary unit is \(i\), defined as \(i = \sqrt{-1}\).

When encountering the square root of a negative number, such as \(\sqrt{-4}\) in the quadratic formula, we convert this into an imaginary number by factoring out \(i\) and then simplifying the non-negative part.

• For \(\sqrt{-4}\), break it down to \(\sqrt{4} \times i\), which becomes \(2i\).

Imaginary numbers are essential in providing solutions to equations that do not have real solutions. In more complex mathematical applications, imaginary numbers combine with real numbers to form complex numbers, expanding the possibilities for solving a variety of real-world problems.