Solving Quadratic Equations A Step-by-Step Guide To Finding Roots
Hey guys! Let's dive into the fascinating world of quadratic equations. Ever wondered how to find the solutions, or roots, of equations like 16x² - 4x - 2 = 0? Well, you’ve come to the right place! In this guide, we'll break down the process step-by-step, making it super easy to understand. We’ll tackle four different equations, so you’ll have a solid grasp of how to solve these problems. So, let's get started and become quadratic equation pros!
Understanding Quadratic Equations
Before we jump into solving specific equations, let's make sure we're all on the same page. A quadratic equation is basically a polynomial equation of the second degree. This means it has the general form: ax² + bx + c = 0, where 'a', 'b', and 'c' are constants, and 'x' is the variable we want to solve for. The key here is the x² term, which makes it quadratic. If you’ve ever wondered why these equations are so important, it’s because they pop up everywhere – from physics problems to engineering calculations, and even in computer graphics! Think about the trajectory of a ball thrown in the air – that's a parabola, which is described by a quadratic equation. Understanding how to solve these equations opens up a whole new world of problem-solving abilities.
Now, why do we care about the solutions, or roots, of these equations? Well, the roots tell us where the parabola (the graph of the quadratic equation) intersects the x-axis. These points are crucial in many applications. For example, in engineering, you might need to find the points where a structural element experiences maximum stress, and these points could be the roots of a quadratic equation. So, finding the roots is not just an abstract math exercise; it has real-world implications. We’ll be using a few different methods to find these roots, including factoring, completing the square, and the quadratic formula. Each method has its strengths and weaknesses, and choosing the right one can make the process much simpler. Stick with us, and you’ll become a master at picking the best approach for any quadratic equation that comes your way!
a) Solving 16x² - 4x - 2 = 0
Let's kick things off with our first equation: 16x² - 4x - 2 = 0. This looks a bit intimidating at first, but don't worry, we'll break it down into manageable steps. The first thing we should always do is check if we can simplify the equation. In this case, we can divide all the terms by 2, which makes our lives much easier. So, dividing by 2, we get:
8x² - 2x - 1 = 0
Now, this looks much cleaner! There are a couple of ways we can tackle this. We could try factoring, which involves finding two binomials that multiply to give us our quadratic. However, factoring isn't always straightforward, especially when the coefficients are a bit tricky. So, let’s go with a more reliable method: the quadratic formula. This formula is a lifesaver and works for any quadratic equation. Remember the general form ax² + bx + c = 0? Our quadratic formula is:
x = [-b ± √(b² - 4ac)] / (2a)
Here, 'a' is the coefficient of x², 'b' is the coefficient of x, and 'c' is the constant term. In our simplified equation, 8x² - 2x - 1 = 0, we have a = 8, b = -2, and c = -1. Now, we just plug these values into the formula:
x = [-(-2) ± √((-2)² - 4 * 8 * -1)] / (2 * 8)
Let's simplify this step by step. First, we deal with the negative signs and the square:
x = [2 ± √(4 + 32)] / 16
Next, we add the numbers under the square root:
x = [2 ± √36] / 16
Ah, √36 is a perfect square, which is great! So, we have:
x = [2 ± 6] / 16
Now, we have two possible solutions, one with the plus sign and one with the minus sign:
x₁ = (2 + 6) / 16 = 8 / 16 = 1/2
x₂ = (2 - 6) / 16 = -4 / 16 = -1/4
So, the solution set for this equation is {1/2, -1/4}. Awesome! We’ve found the roots using the quadratic formula. This method is super powerful because it works every time, even when factoring seems impossible. Remember, the key is to identify 'a', 'b', and 'c' correctly and then carefully plug them into the formula. Keep practicing, and you’ll become a pro at using the quadratic formula!
b) Solving 81x² - 9x - 6 = 0
Alright, let's move on to our second equation: 81x² - 9x - 6 = 0. Just like before, our first step is to see if we can simplify this equation. Notice that 81, 9, and 6 are all divisible by 3. Let’s divide the entire equation by 3 to make it easier to work with. This gives us:
27x² - 3x - 2 = 0
Much better! Now, let's think about our options for solving this quadratic equation. Factoring might be a possibility, but with these coefficients, it could be a bit tricky. To avoid the guesswork, let’s stick with the quadratic formula. It’s reliable and will always get us to the solution. Remember, the quadratic formula is:
x = [-b ± √(b² - 4ac)] / (2a)
In our simplified equation, 27x² - 3x - 2 = 0, we can identify our coefficients: a = 27, b = -3, and c = -2. Now, let's plug these values into the formula:
x = [-(-3) ± √((-3)² - 4 * 27 * -2)] / (2 * 27)
Time to simplify! First, let's deal with the negative signs and the square:
x = [3 ± √(9 + 216)] / 54
Next, we add the numbers under the square root:
x = [3 ± √225] / 54
Great news! √225 is a perfect square, which is 15. So we have:
x = [3 ± 15] / 54
Now we have two potential solutions, one with the plus sign and one with the minus sign:
x₁ = (3 + 15) / 54 = 18 / 54 = 1/3
x₂ = (3 - 15) / 54 = -12 / 54 = -2/9
So, the solution set for this equation is {1/3, -2/9}. Fantastic! We've successfully found the roots using the quadratic formula once again. This reinforces how versatile and dependable the quadratic formula is. Remember, the key is careful substitution and simplification. Keep practicing, and you'll become more and more comfortable with this method!
c) Solving 2x + 3 = (2x + 2)²
Okay, guys, let's tackle our third equation: 2x + 3 = (2x + 2)². This one looks a bit different because it's not in the standard quadratic form right away. Our first job is to get it into the form ax² + bx + c = 0. To do that, we need to expand the right side of the equation. Remember, (2x + 2)² means (2x + 2) * (2x + 2). Let's use the FOIL method (First, Outer, Inner, Last) to expand it:
(2x + 2)² = (2x * 2x) + (2x * 2) + (2 * 2x) + (2 * 2) = 4x² + 4x + 4x + 4 = 4x² + 8x + 4
Now we can rewrite our original equation as:
2x + 3 = 4x² + 8x + 4
To get everything on one side, we’ll subtract 2x and 3 from both sides:
0 = 4x² + 8x + 4 - 2x - 3
Combine like terms:
0 = 4x² + 6x + 1
Great! Now we have a quadratic equation in the standard form: 4x² + 6x + 1 = 0. We can now identify our coefficients: a = 4, b = 6, and c = 1. Let's use the ever-reliable quadratic formula to find the solutions:
x = [-b ± √(b² - 4ac)] / (2a)
Plug in our values:
x = [-6 ± √(6² - 4 * 4 * 1)] / (2 * 4)
Simplify step by step:
x = [-6 ± √(36 - 16)] / 8
x = [-6 ± √20] / 8
√20 isn't a perfect square, but we can simplify it. √20 is the same as √(4 * 5), which is 2√5. So we have:
x = [-6 ± 2√5] / 8
We can simplify this further by dividing all terms by 2:
x = [-3 ± √5] / 4
So, our two solutions are:
x₁ = (-3 + √5) / 4
x₂ = (-3 - √5) / 4
Therefore, the solution set for this equation is {(-3 + √5) / 4, (-3 - √5) / 4}. Awesome! This equation showed us that sometimes we need to do a little algebraic manipulation to get the quadratic equation into the standard form before we can apply the quadratic formula. Keep an eye out for these kinds of equations, and remember to simplify as much as possible!
d) Solving 4x² - (2 + √2) - 2x + 202 = 0
Last but not least, let's tackle our final equation: 4x² - (2 + √2) - 2x + 202 = 0. Whoa, this one looks like a beast! But don't worry, we'll tame it together. The first thing we need to do is make sure the equation is in the standard quadratic form: ax² + bx + c = 0. Let’s rewrite the equation, grouping the terms properly:
4x² - 2x + [202 - (2 + √2)] = 0
Now, let’s simplify the constant term, 'c':
c = 202 - (2 + √2) = 202 - 2 - √2 = 200 - √2
So, our equation in standard form is:
4x² - 2x + (200 - √2) = 0
Now we can identify our coefficients: a = 4, b = -2, and c = 200 - √2. This time, our 'c' is a bit more complex, but that's okay! We're not intimidated. Let's bring out our trusty quadratic formula:
x = [-b ± √(b² - 4ac)] / (2a)
Plug in the values:
x = [-(-2) ± √((-2)² - 4 * 4 * (200 - √2))] / (2 * 4)
Time for some careful simplification! First, deal with the negatives and the square:
x = [2 ± √(4 - 16 * (200 - √2))] / 8
Now, let’s distribute the -16 inside the parentheses:
x = [2 ± √(4 - 3200 + 16√2)] / 8
Combine the numbers under the square root:
x = [2 ± √(-3196 + 16√2)] / 8
Okay, this is interesting. We have a negative number under the square root: -3196 + 16√2. Let’s approximate 16√2. √2 is approximately 1.414, so 16√2 is about 16 * 1.414, which is roughly 22.624. So, we have:
-3196 + 22.624 ≈ -3173.376
Since we have a negative number under the square root, this means we will have complex solutions. In the context of the question, which asks for solutions in the real number universe (R), there are no real solutions for this equation. The discriminant (the value under the square root) is negative, indicating complex roots.
So, the solution set for this equation in the real number universe is ∅ (the empty set). Wow! This equation taught us that not all quadratic equations have real number solutions. Sometimes, the solutions are complex numbers, which are a whole other fascinating area of mathematics! Remember to always check the discriminant to determine the nature of the roots.
Conclusion
And there you have it, guys! We’ve walked through solving four different quadratic equations, each with its own little twist. We’ve seen how to simplify equations, how to use the quadratic formula, and even how to recognize when an equation has no real solutions. Remember, the quadratic formula is your best friend when it comes to solving these equations, but don't forget to look for opportunities to simplify first. Keep practicing, and you’ll master the art of solving quadratic equations in no time! Happy solving!
