Math Problem Statement

Solve each quadratic equation using the quadratic formula, where each equation may involve complex roots.

Solution

The worksheet you've provided contains quadratic equations that need to be solved using the quadratic formula. I'll solve the first few equations as examples, showing the steps in detail.

The quadratic formula is: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} In this formula:

  • aa is the coefficient of x2x^2,
  • bb is the coefficient of xx, and
  • cc is the constant term.

Let's go through some of the problems in detail.

Problem 1

v2+6v+12=0v^2 + 6v + 12 = 0 Here, a=1a = 1, b=6b = 6, and c=12c = 12.

Step 1: Calculate the Discriminant

b24ac=624112=3648=12b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 12 = 36 - 48 = -12 Since the discriminant is negative, the roots will be complex.

Step 2: Apply the Quadratic Formula

v=6±1221=6±12i2=6±23i2=3±3iv = \frac{-6 \pm \sqrt{-12}}{2 \cdot 1} = \frac{-6 \pm \sqrt{12}i}{2} = \frac{-6 \pm 2\sqrt{3}i}{2} = -3 \pm \sqrt{3}i

Solution for Problem 1: v=3±3iv = -3 \pm \sqrt{3}i


Problem 2

3m2+3m+12=03m^2 + 3m + 12 = 0 Here, a=3a = 3, b=3b = 3, and c=12c = 12.

Step 1: Calculate the Discriminant

b24ac=324312=9144=135b^2 - 4ac = 3^2 - 4 \cdot 3 \cdot 12 = 9 - 144 = -135 The discriminant is negative, indicating complex roots.

Step 2: Apply the Quadratic Formula

m=3±13523=3±135i6=3±315i6=12±152im = \frac{-3 \pm \sqrt{-135}}{2 \cdot 3} = \frac{-3 \pm \sqrt{135}i}{6} = \frac{-3 \pm 3\sqrt{15}i}{6} = -\frac{1}{2} \pm \frac{\sqrt{15}}{2}i

Solution for Problem 2: m=12±152im = -\frac{1}{2} \pm \frac{\sqrt{15}}{2}i


Problem 3

b2+6b27=0b^2 + 6b - 27 = 0 Here, a=1a = 1, b=6b = 6, and c=27c = -27.

Step 1: Calculate the Discriminant

b24ac=6241(27)=36+108=144b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-27) = 36 + 108 = 144 The discriminant is positive, so the roots will be real.

Step 2: Apply the Quadratic Formula

b=6±14421=6±122b = \frac{-6 \pm \sqrt{144}}{2 \cdot 1} = \frac{-6 \pm 12}{2} This gives two roots:

  1. b=6+122=3b = \frac{-6 + 12}{2} = 3
  2. b=6122=9b = \frac{-6 - 12}{2} = -9

Solution for Problem 3: b=3b = 3 or b=9b = -9


Would you like solutions for the remaining problems, or do you have specific ones you’d like to focus on?


Further Questions to Explore:

  1. How do complex roots affect the nature of the graph of a quadratic equation?
  2. What happens to the quadratic formula when the discriminant is zero?
  3. Can you use completing the square to verify one of the solutions above?
  4. How does the coefficient aa in ax2+bx+c=0ax^2 + bx + c = 0 influence the solution?
  5. How are complex roots represented in terms of their real and imaginary components?

Tip: Always check the discriminant first; it tells you whether the roots are real or complex before fully solving.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Complex Numbers

Formulas

Quadratic formula x = (-b ± √(b² - 4ac)) / 2a

Theorems

Quadratic formula theorem

Suitable Grade Level

Grades 9-11