Problem 39 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{2}-6 x+13=0 $$

Step-by-Step Solution

Answer

The equation has complex solutions: \(x = 3 \pm 2i\).

Step 1: Identify the Type of Quadratic Equation

The given equation is \(x^{2}-6x+13=0\), which is a quadratic equation of the form \(ax^2 + bx + c = 0\) where \(a=1\), \(b=-6\), and \(c=13\).

Step 2: Calculate the Discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\) to get: \(D = (-6)^2 - 4 \cdot 1 \cdot 13 = 36 - 52 = -16\). This indicates that the equation has complex roots since \(D < 0\).

Step 3: Use the Quadratic Formula

Since the discriminant is negative, use the quadratic formula \(x = \frac{-b \pm \sqrt{D}}{2a}\) to find the complex roots. Substitute \(a=1\), \(b=-6\), and \(D=-16\):\[x = \frac{-(-6) \pm \sqrt{-16}}{2 \cdot 1} = \frac{6 \pm 4i}{2}\].

Step 4: Simplify the Expression

Simplify the expression for the roots:\[x = \frac{6}{2} \pm \frac{4i}{2} = 3 \pm 2i\].

Step 5: Conclude the Solution

The solutions to the equation are \(x = 3 + 2i\) and \(x = 3 - 2i\). These are complex conjugates of each other.

Key Concepts

Complex Roots

When solving quadratic equations, complex roots appear when the discriminant is negative. But what are complex roots? Simply put, complex numbers are numbers that have a real part and an imaginary part. The imaginary unit, represented as "i," is defined as \( \sqrt{-1} \).
When a quadratic equation does not have real solutions, it means its graph does not intersect the x-axis. Instead, the solutions are not ordinary real numbers, but complex numbers.
In the example equation \( x^{2} - 6x + 13 = 0 \), we determined that the solutions are \( x = 3 + 2i \) and \( x = 3 - 2i \). Notice how these solutions are complex conjugates. Complex conjugates come in pairs, forming a mirror image about the horizontal axis in the complex plane. If your equation has real coefficients, the complex roots will always occur in these conjugate pairs.

Discriminant

The discriminant is a valuable part of solving a quadratic equation since it tells us about the nature of the roots. You get the discriminant from the standard form \( ax^2 + bx + c = 0 \) by the formula \( D = b^2 - 4ac \).
Here's how to interpret the discriminant:

• If \( D > 0 \), there are two distinct real roots.
• If \( D = 0 \), there is exactly one real root, and it is repeated.
• If \( D < 0 \), as in our original exercise where \( D = -16 \), the roots are complex conjugates, meaning they contain the imaginary unit "i."

By carefully evaluating the discriminant, you can predict the number and type of solutions without fully solving the equation. This makes it a powerful tool in mathematics.

Quadratic Formula

The quadratic formula is a universal method to solve any quadratic equation. It's expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula helps us find the roots by plugging in the values of \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).
In our case, with \( a=1 \), \( b=-6 \), and \( c=13 \), we substitute these values, and since \( D = -16 \), it results in a complex solution. The term \( \sqrt{-16} \) converts to \( \pm 4i \). Once plugged in, the equation simplifies to \( x = 3 \pm 2i \).
The quadratic formula is especially useful when factoring is not obvious or when dealing with complex roots.

Completing the Square

Another approach to solving quadratic equations is completing the square. This method involves transforming the quadratic equation into a perfect square trinomial.
Let's apply it to our original exercise, \( x^{2} - 6x + 13 = 0 \). Start with the equation:\[x^2 - 6x = -13\]Find a number that makes the left side a perfect square trinomial. Take half of the coefficient of \( x \) (which is \(-6\)), divide it by 2 to get \(-3\), and then square it to get \(9\).
Add and subtract this inside the equation:\[x^2 - 6x + 9 = 9 - 13\]Rewrite it as:\[(x - 3)^2 = -4\]Solve for \(x\) by taking the square root of both sides:\(x - 3 = \pm 2i\). This method confirms the solution \( x = 3 \pm 2i \) by showing clearly how \(x\) evolves from a real square to incorporating imaginary parts.