Problem 23 - Algebra A Combined Function

Question

Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. $$ x^{2}+6 x+13=0 $$

Step-by-Step Solution

Answer

The solutions are \(-3 + 2i\) and \(-3 - 2i\).

Step 1: Identify Coefficients

To solve the quadratic equation using the quadratic formula, we first identify the coefficients from the equation \(x^2 + 6x + 13 = 0\). Here, \(a = 1\), \(b = 6\), and \(c = 13\).

Step 2: Write the Quadratic Formula

The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] We will use this formula to find the values of \(x\).

Step 3: Calculate the Discriminant

The discriminant \(D\) is calculated as \(b^2 - 4ac\). Substitute the values we have: \[ D = 6^2 - 4(1)(13) = 36 - 52 = -16 \] Since the discriminant is negative, the solutions will be complex numbers.

Step 4: Substitute into Quadratic Formula

Substituting the values of \(a\), \(b\), and the discriminant into the quadratic formula gives us: \[ x = \frac{-6 \pm \sqrt{-16}}{2(1)} \]

Step 5: Simplify the Square Root of the Discriminant

The square root of \(-16\) simplifies to \(4i\) because \(\sqrt{-16} = \sqrt{-1} \times \sqrt{16} = 4i\). So, we have: \[ x = \frac{-6 \pm 4i}{2} \]

Step 6: Divide and Simplify

Now we divide each term by 2 to simplify the expression: \[ x = \frac{-6}{2} \pm \frac{4i}{2} = -3 \pm 2i \]So, the solutions are \(-3 + 2i\) and \(-3 - 2i\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). This formula provides a straightforward way to find the roots of the equation, even when the solutions are complex numbers. The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

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

To use this formula, a few simple steps need to be followed. First, identify the coefficients \(a, b,\) and \(c\) from your quadratic equation. Second, calculate the discriminant \(b^2 - 4ac\). Finally, substitute all these values into the formula to find the solutions for \(x\). These solutions might be real numbers or complex numbers depending on the discriminant.

Complex Numbers

Complex numbers are numbers that include a real part and an imaginary part. They are usually written in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part multiplied by the unit imaginary number \(i\), which is defined as \(\sqrt{-1}\). The beauty of complex numbers is that they expand the range of solvable quadratic equations.In quadratic equations where the discriminant is negative, the solutions will involve complex numbers since taking the square root of a negative number results in a combination of real and imaginary numbers. For example, solving the equation \(x^2 + 6x + 13 = 0\) gives us the solutions \(-3 + 2i\) and \(-3 - 2i\). Here, the number \(-3\) is the real part while \(2i\) represents the imaginary part. Complex numbers are crucial in representing solutions that can't be solved on the traditional real number line. They find applications in engineering, physics, and applied mathematics, spreading far beyond theoretical mathematics.

Discriminant

The discriminant in a quadratic equation is a key part of determining the nature of the roots of the equation. Denoted as \(b^2 - 4ac\), the discriminant tells us whether the roots are real or complex.

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it equals zero, the equation has exactly one real root (or a repeated root).
• If it is negative, as in our example \(-16\), the equation has two complex roots.

Calculating the discriminant is a straightforward process. For the equation \(x^2 + 6x + 13 = 0\), the discriminant calculation goes as follows:\[ D = 6^2 - 4 \times 1 \times 13 = 36 - 52 = -16 \]The negative discriminant in this example indicates that the solutions are complex. Understanding the discriminant helps students anticipate the nature of the roots before solving the equation, making the process intuitive and logical.