Solve This Quadratic Equation: 4x² - 30x + 45 = 0

Hey guys, let's dive into solving a quadratic equation today! Specifically, we're tackling the beast: $4 x^2-30 x+45=0$. Now, I know quadratic equations can sometimes look a bit intimidating, but trust me, with the right tools and a little practice, you'll be acing them in no time. We're going to break down how to find those elusive solutions, often called roots, and figure out which of the provided options is the correct one. So, grab your notebooks, and let's get this math party started!

Understanding Quadratic Equations

Alright, so what exactly is a quadratic equation? In its most basic form, it's an equation that can be written as $ax^2 + bx + c = 0$, where 'a', 'b', and 'c' are numbers, and crucially, 'a' cannot be zero. If 'a' were zero, it would just be a linear equation, which is a whole different ballgame. The 'x²' term is what gives it the 'quadratic' name – it's the highest power of our variable, 'x'. Finding the solutions to a quadratic equation means finding the values of 'x' that make the equation true. These solutions can be real numbers, complex numbers, or sometimes there might be only one solution (a repeated root), or even no real solutions at all. It all depends on the numbers 'a', 'b', and 'c'. For our specific equation, $4 x^2-30 x+45=0$, we can see that $a=4$, $b=-30$, and $c=45$. These are the key players we'll be working with.

There are a few common methods to solve quadratic equations. You've got factoring, completing the square, and the ever-reliable quadratic formula. Factoring is great when it works, but it's not always straightforward, especially with larger numbers. Completing the square is a solid method and is actually how the quadratic formula is derived, but it can be a bit tedious. That's why, for a general quadratic equation like ours where factoring might not be obvious, the quadratic formula is usually our go-to hero. It's a universal key that unlocks the solutions for any quadratic equation. It looks a bit daunting at first glance, but once you understand it and practice plugging in the numbers, it becomes your best friend. So, before we jump into solving our specific problem, let's quickly recap the quadratic formula itself. Remember it, write it down, tattoo it on your brain – whatever it takes!

The Mighty Quadratic Formula

The quadratic formula is the golden ticket to solving any equation in the form $ax^2 + bx + c = 0$. It states that the solutions for 'x' are given by:

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

See that $\pm$ symbol? That's super important! It means there are potentially two solutions: one where you add the square root part, and one where you subtract it. The part under the square root, $b^2 - 4ac$, is called the discriminant. This little guy tells us a lot about the nature of the solutions before we even calculate them. If the discriminant is positive, we get two distinct real solutions. If it's zero, we get exactly one real solution (a repeated root). And if it's negative, well, that means we're dealing with complex solutions involving the imaginary unit 'i'. For our equation, $4 x^2-30 x+45=0$, we've identified $a=4$, $b=-30$, and $c=45$. Now, it's time to plug these values into the quadratic formula and see what magic happens.

Remember, the accuracy of your final answer hinges entirely on how carefully you substitute these values and perform the arithmetic. Mistakes often happen in the sign changes or when squaring negative numbers. So, double-checking each step is crucial. It might seem like a lot of work, but think of it as a detective process – gathering clues (the values of a, b, and c) and applying a known method (the quadratic formula) to find the truth (the solutions for x). We're not just blindly plugging numbers in; we're systematically solving a mathematical puzzle. Let's make sure we handle the negative sign of 'b' correctly when we substitute it into the formula, as this is a common pitfall. Also, be mindful of the order of operations, especially within the square root. Squaring 'b', then multiplying $4ac$, and then subtracting $4ac$ from $b^2$ needs to be done precisely. The division by $2a$ comes last. We'll walk through each of these steps methodically to ensure we get the correct result. Don't rush it; precision is key in mathematics!

Plugging in the Values

Okay, team, let's get our hands dirty with the actual calculation. We have our equation $4 x^2-30 x+45=0$, so we know:

• $a = 4$
• $b = -30$
• $c = 45$

Now, we substitute these values into the quadratic formula:

$$x = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(4)(45)}}{2(4)}$$

See how we replaced 'b' with '-30'? That double negative, $-(-30)$, is going to become a positive 30. This is a super common place to trip up, so always pay close attention to those signs!

Let's simplify the components step-by-step. First, the numerator:

• $-(-30) = 30$
• $(-30)^2 = 900$ (Remember, a negative number squared is always positive!)
• $4(4)(45) = 16(45)$. To calculate $16 imes 45$: $16 imes 40 = 640$, and $16 imes 5 = 80$. So, $640 + 80 = 720$.

Now, let's put that back into the square root (the discriminant):

• $b^2 - 4ac = 900 - 720 = 180$

So, our formula now looks like this:

$$x = \frac{30 \pm \sqrt{180}}{2(4)}$$

And the denominator is easy:

• $2(4) = 8$

Giving us:

$$x = \frac{30 \pm \sqrt{180}}{8}$$

We're getting closer, guys! The next crucial step is simplifying that square root, $\sqrt{180}$.

When simplifying radicals like $\sqrt{180}$, we're looking for perfect square factors within 180. Think of perfect squares: 4, 9, 16, 25, 36, 49, etc. Let's see if any of these divide evenly into 180.

• 180 divided by 4 is 45. So, $\sqrt{180} = \sqrt{4 \times 45} = \sqrt{4} \times \sqrt{45} = 2\sqrt{45}$. This is better, but we can simplify $\sqrt{45}$ further.
• 45 has a factor of 9 (which is a perfect square!). $45 = 9 imes 5$. So, $\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$.

Putting it all together: $\sqrt{180} = 2\sqrt{45} = 2(3\sqrt{5}) = 6\sqrt{5}$.

Alternatively, you could recognize that $180 = 36 imes 5$. Since 36 is a perfect square ($6^2$), $\sqrt{180} = \sqrt{36 imes 5} = \sqrt{36} \times \sqrt{5} = 6\sqrt{5}$. This is the quickest way if you spot the largest perfect square factor. Simplifying radicals is a skill that gets easier with practice, so don't get discouraged if it takes a moment!

Now, let's substitute this simplified radical back into our equation for x:

$$x = \frac{30 \pm 6\sqrt{5}}{8}$$

We're almost there! One final simplification step often involves dividing all the terms in the numerator and the denominator by their greatest common divisor. In this case, 30, 6, and 8 are all divisible by 2.

• $30 \div 2 = 15$
• $6\sqrt{5} \div 2 = 3\sqrt{5}$
• $8 \div 2 = 4$

So, our simplified solutions are:

$$x = \frac{15 \pm 3\sqrt{5}}{4}$$

This looks like one of our answer choices, doesn't it? Let's take a peek at the options provided to confirm.

Checking the Answer Choices

We've worked hard to arrive at our solution: $x = \frac{15 \pm 3\sqrt{5}}{4}$. Now, let's compare this to the options given:

A. $\frac{-15 \pm 6 \sqrt{5}}{4}$ B. $\frac{15 \pm 6 \sqrt{5}}{4}$ C. $\frac{15 \pm 3 \sqrt{5}}{4}$ D. $\frac{30 \pm 3 \sqrt{5}}{4}$

Boom! Option C matches our calculated solution perfectly. The '$\\pm$' symbol indicates the two distinct roots: $\frac{15 + 3\sqrt{5}}{4}$ and $\frac{15 - 3\sqrt{5}}{4}$. Option A has the wrong sign for the '15'. Option B has the coefficient of $\sqrt{5}$ incorrect (it should be 3, not 6). Option D has the numerator terms incorrect before simplification. So, we can confidently say that C is the correct answer.

Remember, checking your work is always a good idea. If you have time, you could even plug these values back into the original equation $4 x^2-30 x+45=0$ to verify they make the equation true. However, given the complexity of the numbers, simply comparing your derived formula to the choices is often sufficient, especially if you've been meticulous with your calculations. The process of simplifying the radical and then reducing the fraction are key steps where errors can creep in, so reviewing those parts of your calculation is wise.

Why Quadratic Formulas Matter

So, why do we bother learning about quadratic equations and their solutions? Well, these aren't just abstract math problems cooked up to torture students (though sometimes it feels like it, right?). Quadratic equations pop up everywhere in the real world. Think about projectile motion – like the path of a ball thrown in the air. That parabolic arc is described by a quadratic equation! Engineers use them to design bridges and buildings, ensuring structural integrity. Physicists use them to model everything from the movement of planets to the behavior of subatomic particles. Economists use them for cost analysis and profit maximization. Even in computer graphics, they're used to create smooth curves and animations.

Understanding how to solve them gives you a powerful toolset for analyzing and predicting phenomena in a vast range of fields. It builds your logical thinking and problem-solving skills, which are valuable no matter what career path you choose. The ability to break down a complex problem into smaller, manageable steps, identify the correct tools (like the quadratic formula), and execute the solution systematically is a transferable skill that employers highly value. It's about more than just getting the right number; it's about developing a rigorous and analytical mindset. So, the next time you're wrestling with a quadratic equation, remember you're honing skills that are applicable far beyond the classroom. You're becoming a better problem-solver, and that's a win in my book!

Keep practicing, guys! The more you solve these, the more comfortable and confident you'll become. Don't be afraid to make mistakes; they're just stepping stones to understanding. Keep that math spirit alive!