Problem 37 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ r^{2}-6 r+14=0 $$

Step-by-Step Solution

Answer

The solutions are \( 3 + \sqrt{5}i \) and \( 3 - \sqrt{5}i \).

Step 1: Identify coefficients

Identify the coefficients in the quadratic equation. For the equation \( r^{2} - 6r + 14 = 0 \), the coefficients are: \( a = 1 \), \( b = -6 \), and \( c = 14 \).

Step 2: Write the quadratic formula

Write down the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]

Step 3: Calculate the discriminant

Calculate the discriminant using the formula \( b^{2} - 4ac \). For our coefficients, it is: \[ (-6)^{2} - 4 \cdot 1 \cdot 14 = 36 - 56 = -20 \] Since the discriminant is negative, the equation has non-real complex solutions.

Step 4: Substitute into the quadratic formula

Substitute the coefficients and the discriminant into the quadratic formula: \[ r = \frac{-(-6) \pm \sqrt{-20}}{2 \cdot 1} = \frac{6 \pm \sqrt{-20}}{2} \]

Step 5: Simplify the expression

Simplify the expression by breaking down the square root of the negative discriminant: \[ \sqrt{-20} = \sqrt{20}i = 2\sqrt{5}i \] Then substitute back into the equation: \[ r = \frac{6 \pm 2\sqrt{5}i}{2} = 3 \pm \sqrt{5}i \]

Key Concepts

Quadratic Formula

To solve quadratic equations like \(r^2 - 6r + 14 = 0\), we use the quadratic formula. This formula is a key tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). The quadratic formula is:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Let's break it down step-by-step:

• \(a\) is the coefficient of \(r^2\)
• \(b\) is the coefficient of \(r\)
• \(c\) is the constant term

This formula calculates the values of \(r\) that satisfy the equation. You’ll often use it in algebra to find solutions to various quadratic equations.

Complex Numbers

When solving a quadratic equation you might find solutions in the form of complex numbers. A complex number is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, defined as \(\sqrt{-1}\).
Complex numbers are useful when the solutions to an equation involve the square root of a negative number.

• Real part: \(a\)
• Imaginary part: \(bi\)

In our example, the discriminant is negative, leading us to complex solutions:
\(3 \pm \sqrt{5}i\). This means the solutions are non-real but still valid for the quadratic equation.

Discriminant Calculation

The discriminant helps determine the nature of the solutions for a quadratic equation. It is the part of the quadratic formula under the square root: \(b^2 - 4ac\). The value of the discriminant can tell us a lot:

• If \(b^2 - 4ac > 0\), we have two real and distinct solutions.
• If \(b^2 - 4ac = 0\), we have exactly one real solution.
• If \(b^2 - 4ac < 0\), we have two complex solutions.

In our problem, the discriminant is:
\[(-6)^2 - 4 \cdot 1 \cdot 14 = 36 - 56 = -20\] Since it is negative, this explains why the solutions are complex numbers.

Non-Real Solutions

Non-real solutions occur when the discriminant of a quadratic equation is negative. These solutions involve complex numbers, meaning they have both real and imaginary parts. For our equation \(r^2 - 6r + 14 = 0\), we calculated a negative discriminant, which resulted in non-real solutions.
After substituting the discriminant into the quadratic formula and simplifying, we get:
\[ r = 3 \pm \sqrt{5}i \] These solutions, \(3 + \sqrt{5}i\) and \(3 - \sqrt{5}i\), are non-real because they have imaginary parts. Non-real solutions are crucial in many advanced fields of mathematics and physics, though their basics are covered well in algebra classes.