Complex Zeros Of Quadratic Functions

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a problem that might seem a little intimidating at first glance: finding the complex zeros of a quadratic function. Don't worry, we'll break it down step-by-step, making it super clear and easy to understand. So, grab your notebooks and let's get started on unraveling the mysteries of $f(x)=x^2-8x+17$!

Understanding Quadratic Functions and Their Zeros

Before we jump into finding the complex zeros, let's quickly recap what quadratic functions and their zeros are all about. A quadratic function is a polynomial function of degree two, generally expressed in the form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants and $a eq 0$. The zeros of a function, also known as roots, are the values of $x$ for which $f(x) = 0$. In simpler terms, they are the $x$-intercepts of the function's graph on the coordinate plane. For most quadratic functions, we can find these zeros by factoring, completing the square, or using the quadratic formula. However, sometimes the solutions aren't real numbers; they involve the imaginary unit $i$, which is defined as $\sqrt{-1}$. These are what we call complex zeros.

The Quadratic Formula: Your Best Friend for Complex Zeros

When factoring doesn't seem to work, or when we suspect complex zeros might be involved, the quadratic formula is our go-to tool. Remember this beauty? For any quadratic equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

The part under the square root, $b^2 - 4ac$, is called the discriminant. The discriminant tells us a lot about the nature of the roots:

• If $b^2 - 4ac > 0$, there are two distinct real roots.
• If $b^2 - 4ac = 0$, there is exactly one real root (a repeated root).
• If $b^2 - 4ac < 0$, there are two distinct complex roots (a conjugate pair).

In our case, the function is $f(x) = x^2 - 8x + 17$. To find the zeros, we set $f(x) = 0$, so we have the equation $x^2 - 8x + 17 = 0$. Here, we can identify our coefficients: $a = 1$, $b = -8$, and $c = 17$. Let's plug these values into the quadratic formula and see what we get.

Solving $f(x)=x^2-8x+17$ for Complex Zeros

Alright guys, let's get down to business with our specific function, $f(x) = x^2 - 8x + 17$. Our mission is to find the values of $x$ that make $f(x) = 0$. So, we're solving the equation $x^2 - 8x + 17 = 0$. First off, let's check the discriminant to see what kind of roots we're dealing with. Remember, the discriminant is $b^2 - 4ac$. Plugging in our values $a=1$, $b=-8$, and $c=17$, we get:

$$\Delta = (-8)^2 - 4(1)(17)$$

$$\Delta = 64 - 68$$

$$\Delta = -4$$

Uh oh! The discriminant is negative ($-4$). This tells us immediately that we're going to have complex zeros. This is where the fun begins, because we get to work with imaginary numbers! The quadratic formula will guide us through this, and it's pretty straightforward once you know the drill. The formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. We already calculated the discriminant ($b^2 - 4ac = -4$), so we can substitute that in.

Applying the Quadratic Formula Step-by-Step

Let's continue with the quadratic formula: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$. We have $a=1$, $b=-8$, and $\Delta = -4$. So, substituting these values:

$$x = \frac{-(-8) \pm \sqrt{-4}}{2(1)}$$

$$x = \frac{8 \pm \sqrt{-4}}{2}$$

Now, we need to deal with the square root of $-4$. Remember our good friend, the imaginary unit $i$, where $i = \sqrt{-1}$? We can rewrite $\sqrt{-4}$ as $\sqrt{4 \times -1}$, which is $\sqrt{4} \times \sqrt{-1}$. Since $\sqrt{4} = 2$ and $\sqrt{-1} = i$, we have $\sqrt{-4} = 2i$. Perfect! Let's plug this back into our equation for $x$:

$$x = \frac{8 \pm 2i}{2}$$

To simplify this further, we can divide both terms in the numerator by the denominator (2):

$$x = \frac{8}{2} \pm \frac{2i}{2}$$

$$x = 4 \pm i$$

And there you have it, guys! The complex zeros of the quadratic function $f(x) = x^2 - 8x + 17$ are $x = 4 + i$ and $x = 4 - i$. Notice how they come in a conjugate pair? That's a hallmark of quadratic equations with real coefficients that have complex roots. Pretty neat, right?

Verifying the Complex Zeros

It's always a good idea to double-check our work, especially when dealing with complex numbers. Let's verify if our solutions, $x = 4 + i$ and $x = 4 - i$, actually make $f(x) = 0$. We'll substitute each one back into the original function $f(x) = x^2 - 8x + 17$.

Checking $x = 4 + i$

Let's plug in $x = 4 + i$:

$$f(4+i) = (4+i)^2 - 8(4+i) + 17$$

First, let's expand $(4+i)^2$. Using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$$(4+i)^2 = 4^2 + 2(4)(i) + i^2$$

(4+i)^2 = 16 + 8i + (-1)$ (Remember, $i^2 = -1$) $(4+i)^2 = 15 + 8i

Now, let's distribute the $-8$ in the second term:

$$-8(4+i) = -32 - 8i$$

Putting it all together:

$$f(4+i) = (15 + 8i) + (-32 - 8i) + 17$$

Group the real and imaginary parts:

$$f(4+i) = (15 - 32 + 17) + (8i - 8i)$$

$$f(4+i) = (32 - 32) + (0i)$$

$$f(4+i) = 0$$

Awesome! It works for $x = 4 + i$.

Checking $x = 4 - i$

Now let's check the other zero, $x = 4 - i$:

$$f(4-i) = (4-i)^2 - 8(4-i) + 17$$

Expand $(4-i)^2$. Using the formula $(a-b)^2 = a^2 - 2ab + b^2$:

$$(4-i)^2 = 4^2 - 2(4)(i) + i^2$$

$$(4-i)^2 = 16 - 8i + (-1)$$

$$(4-i)^2 = 15 - 8i$$

Distribute the $-8$:

$$-8(4-i) = -32 + 8i$$

Putting it all together:

$$f(4-i) = (15 - 8i) + (-32 + 8i) + 17$$

Group the real and imaginary parts:

$$f(4-i) = (15 - 32 + 17) + (-8i + 8i)$$

$$f(4-i) = (32 - 32) + (0i)$$

$$f(4-i) = 0$$

Both complex zeros, $4+i$ and $4-i$, successfully make the function equal to zero. This confirms our calculations are correct, guys! This verification step is super important to ensure you haven't made any silly arithmetic errors along the way.

Why Complex Zeros Matter

So, why do we even bother with complex zeros? It might seem abstract, but understanding complex numbers and their role in solving equations is fundamental in many areas of science, engineering, and advanced mathematics. For instance, in electrical engineering, AC circuits are analyzed using complex numbers to represent voltage and current. In quantum mechanics, complex numbers are essential for describing wave functions. Even in signal processing and control theory, complex numbers help model and analyze system behavior.

For quadratic functions specifically, the existence of complex zeros means that the parabola represented by the function does not intersect the x-axis. Instead, it