Solving $8x^2 - 10x = -9$: Find The Roots

Hey math whizzes! Ever stared at a quadratic equation and wondered, "What in the world are the solutions to this thing?" Well, today, guys, we're diving deep into one such equation: $8x^2 - 10x = -9$. We're going to break it down, step-by-step, so you can confidently tackle these kinds of problems. Forget those confusing formulas for a sec; let's make sense of it together!

Understanding Quadratic Equations: The Basics You Need to Know

Alright, before we jump into solving our specific problem, let's get our heads around what a quadratic equation actually is. Simply put, a quadratic equation is a polynomial equation of the second degree. This means it has at least one term that is squared. The general form you'll often see is $ax^2 + bx + c = 0$, where 'a', 'b', and 'c' are constants, and 'a' can't be zero (because if it were, it wouldn't be quadratic anymore, right?). Finding the solutions, or roots, of a quadratic equation means finding the values of 'x' that make the equation true. Think of it as finding where the parabola represented by the equation hits the x-axis. Sometimes it hits twice, sometimes once (at the vertex), and sometimes, it doesn't hit the x-axis at all, meaning the solutions are imaginary. The number of solutions can be determined by the discriminant, which is part of the famous quadratic formula. We'll get to that!

Preparing Our Equation for Solution

So, our equation is $8x^2 - 10x = -9$. The first thing we gotta do, folks, is get it into that standard $ax^2 + bx + c = 0$ form. This makes it super easy to identify our 'a', 'b', and 'c' values. To do this, we simply need to move the '-9' from the right side of the equation over to the left side. We do this by adding 9 to both sides: $8x^2 - 10x + 9 = -9 + 9$. This simplifies to $8x^2 - 10x + 9 = 0$. Boom! Now we have it in the standard form. From here, we can clearly see that $a = 8$, $b = -10$, and $c = 9$. Keep these numbers handy, as they're crucial for the next steps.

The Quadratic Formula: Your Best Friend for Solving Quadratics

When faced with a quadratic equation, especially one that doesn't easily factor, the quadratic formula is your golden ticket. It's a universal solution that works for any quadratic equation. The formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Remember this bad boy! It might look a little intimidating at first, but once you plug in your values for 'a', 'b', and 'c', it becomes way less scary. The '±' symbol is super important; it tells us there are potentially two solutions – one where you add the square root part, and one where you subtract it. This is because parabolas can cross the x-axis at two different points.

Plugging in the Values: Let's Do Some Math!

Okay, guys, time for the moment of truth! We've got our equation $8x^2 - 10x + 9 = 0$, so $a = 8$, $b = -10$, and $c = 9$. Let's substitute these into the quadratic formula: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(8)(9)}}{2(8)}$.

First, let's simplify the parts inside the formula. $-(-10)$ becomes $+10$. $(-10)^2$ is $100$. And $4(8)(9)$ is $4 imes 72$, which equals $288$. So, the formula now looks like: $x = \frac{10 \pm \sqrt{100 - 288}}{16}$.

Now, let's tackle the part under the square root: $100 - 288$. This gives us $-188$. Uh oh, a negative number under the square root! This tells us our solutions are going to be imaginary numbers. Remember, the square root of a negative number involves 'i', the imaginary unit, where $i = \sqrt{-1}$. So, $\sqrt{-188}$ can be rewritten as $\sqrt{188} imes \sqrt{-1}$, which is $\sqrt{188}i$. We can simplify $\sqrt{188}$ further. Let's find the largest perfect square that divides 188. $188 = 4 imes 47$. Since 4 is a perfect square, $\sqrt{188} = \sqrt{4 \times 47} = \sqrt{4} \times \sqrt{47} = 2\sqrt{47}$.

So, our square root term becomes $2\sqrt{47}i$. Plugging this back into our formula, we get: $x = \frac{10 \pm 2\sqrt{47}i}{16}$.

Simplifying the Solutions

We're almost there, folks! The last step is to simplify the entire expression. Notice that the numerator ($10$ and $2\sqrt{47}i$) and the denominator ($16$) all share a common factor of 2. Let's divide each term by 2:

$x = \frac{10/2 \pm (2\sqrt{47}i)/2}{16/2}$

$x = \frac{5 \pm \sqrt{47}i}{8}$

This gives us our two solutions:

$x_1 = \frac{5 + i\sqrt{47}}{8}$

$x_2 = \frac{5 - i\sqrt{47}}{8}$

These are the expressions that solve the quadratic equation $8x^2 - 10x = -9$. Pretty neat, right? You've just navigated through solving a quadratic equation with imaginary roots using the trusty quadratic formula!

Comparing Our Solutions to the Options

Now, let's look back at the options you were given:

A. $\frac{-5+\sqrt{97}}{8}$ and $\frac{-5-\sqrt{97}}{8}$ B. $\frac{5+\sqrt{97}}{8}$ and $\frac{5-\sqrt{97}}{8}$ C. $\frac{5+i \sqrt{47}}{8}$ and $\frac{5-i \sqrt{47}}{8}$

By comparing our calculated solutions, $x = \frac{5 \pm i\sqrt{47}}{8}$, with the given options, it's clear that Option C matches our results perfectly. Option A and B would have resulted if the initial equation had different coefficients or if there was a sign error during the calculation, especially when dealing with the discriminant. The presence of the imaginary unit 'i' in Option C confirms that our solutions are indeed complex (imaginary in this case) as indicated by the negative discriminant we found earlier. It's always a good idea to double-check your work, especially when dealing with negative signs and square roots, to ensure accuracy. We've successfully used the quadratic formula to find the roots of the equation $8x^2 - 10x = -9$ and identified the correct answer among the choices.

Final Thoughts on Tackling Quadratics

So there you have it, guys! We've successfully transformed the equation $8x^2 - 10x = -9$ into the standard quadratic form, identified the coefficients, and applied the quadratic formula. We even grappled with imaginary numbers and simplified our final expressions. The key takeaways here are to always put your equation in standard form first, know your 'a', 'b', and 'c' values, and don't be afraid of the quadratic formula – it's your best friend! And remember, when you see a negative under the square root, that's your cue for imaginary solutions involving 'i'. Keep practicing these problems, and you'll become a quadratic equation pro in no time. Math might seem tricky sometimes, but with a little patience and the right tools, you can solve anything. Keep up the awesome work!