Question

question 1

use the quadratic formula to solve the equation (x^2 + 6x + 13 = 0).

enter multiple answers as a list separated by commas. example: (2 + 2i, 2 - 2i)

question 2

solve equation by the quadratic formula. list the solutions, separated by commas.

(2r^2 + 3r + 12 = 8)

(r =)

Explanation:

🆕 New Concept Discovered: Solving Quadratic Equations with Complex Roots
Finding solutions when the discriminant is negative

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Question 1

Solve the equation:

$$ x^2 + 6x + 13 = 0 $$

Step 1: Identify coefficients

Identify the coefficients \( a \), \( b \), and \( c \) from the standard form \( ax^2 + bx + c = 0 \):

$$ a = 1, \quad b = 6, \quad c = 13 $$

Step 2: Apply the quadratic formula

The quadratic formula is:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substitute the coefficients into the formula:

$$ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(13)}}{2(1)} $$

$$ x = \frac{-6 \pm \sqrt{36 - 52}}{2} $$

$$ x = \frac{-6 \pm \sqrt{-16}}{2} $$

Step 3: Simplify using complex numbers

Since we have a negative number under the square root, we use the imaginary unit \( i \) where \( \sqrt{-1} = i \):

$$ \sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i $$

Substitute this back into the expression:

$$ x = \frac{-6 \pm 4i}{2} $$

$$ x = -3 \pm 2i $$

The solutions are:

$$ -3 + 2i, -3 - 2i $$

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Question 2

Solve the equation:

$$ 2r^2 + 3r + 12 = 8 $$

Step 1: Write in standard form

Subtract \( 8 \) from both sides to set the quadratic equation to \( 0 \):

$$ 2r^2 + 3r + 4 = 0 $$

Identify the coefficients:

$$ a = 2, \quad b = 3, \quad c = 4 $$

Step 2: Apply the quadratic formula

Substitute the coefficients into the quadratic formula:

$$ r = \frac{-3 \pm \sqrt{3^2 - 4(2)(4)}}{2(2)} $$

$$ r = \frac{-3 \pm \sqrt{9 - 32}}{4} $$

$$ r = \frac{-3 \pm \sqrt{-23}}{4} $$

Step 3: Simplify using complex numbers

Express the square root of the negative number using \( i \):

$$ \sqrt{-23} = i\sqrt{23} $$

Substitute this back into the expression:

$$ r = \frac{-3 \pm i\sqrt{23}}{4} $$

The solutions are:

$$ -\frac{3}{4} + \frac{\sqrt{23}}{4}i, -\frac{3}{4} - \frac{\sqrt{23}}{4}i $$

(or written as \(\frac{-3 + i\sqrt{23}}{4}, \frac{-3 - i\sqrt{23}}{4}\))

Answer:

Question 1:

$$ -3 + 2i, -3 - 2i $$

Question 2:

$$ \frac{-3 + i\sqrt{23}}{4}, \frac{-3 - i\sqrt{23}}{4} $$