Let's dive into how to solve for x using the quadratic formula. Guys, this is a super important tool in algebra, and once you get the hang of it, you'll be solving quadratic equations like a pro. The quadratic formula is especially useful when you can't easily factor a quadratic equation. It's like your trusty Swiss Army knife for solving these types of problems. So, buckle up, and let's get started!
The quadratic formula is:
x = (-b ± √(b² - 4ac)) / (2a)
This formula helps you find the solutions (also called roots or zeros) of any quadratic equation in the standard form: ax² + bx + c = 0. The a, b, and c are coefficients – just the numbers in front of the variables and the constant term. The ± symbol means you have to calculate two solutions: one with addition and one with subtraction. Remember, paying attention to detail is key here. A small mistake can lead to a wrong answer, so double-check your work as you go. Now, let's jump into our first example.
Example 1: x² – 4 = 5x
Our first equation is x² – 4 = 5x. The first thing we need to do is rearrange it into the standard form ax² + bx + c = 0. To do that, we subtract 5x from both sides of the equation. This gives us:
x² - 5x - 4 = 0
Now, we can identify our coefficients: a = 1, b = -5, and c = -4. Next, we plug these values into the quadratic formula:
x = (-(-5) ± √((-5)² - 4(1)(-4))) / (2(1))
Simplify step by step. First, deal with the negatives and exponents:
x = (5 ± √(25 + 16)) / 2
Then, add the numbers under the square root:
x = (5 ± √41) / 2
So, the two solutions for x are:
x = (5 + √41) / 2
and
x = (5 - √41) / 2
These are the exact solutions. If you need decimal approximations, you can use a calculator to find the square root of 41 and complete the calculations. But for exact solutions, we leave it in this form. See? Not too scary once you break it down. The quadratic formula might seem intimidating at first, but with practice, it becomes second nature. Just remember the steps: rearrange the equation, identify a, b, and c, plug them into the formula, and simplify. And always double-check your signs!
Next up, let's tackle quadratic inequalities and express their solutions in interval notation. This is another crucial skill in algebra, guys. Understanding how to represent solutions using intervals can make a big difference, especially when you move on to more advanced math topics. So, what's the deal with interval notation? It's a way of writing down a set of numbers using intervals, which are basically ranges of values. We use parentheses and brackets to show whether the endpoints of the interval are included or not. This notation is super handy because it provides a concise way to describe the solution set of an inequality. Let's see how it works in practice.
Example 2: x² - 4x - 5 > 0
We have the inequality x² - 4x - 5 > 0. The first step is to factor the quadratic expression. Factoring helps us find the critical points, which are the values of x that make the expression equal to zero. These points divide the number line into intervals, which we then test to see if they satisfy the inequality. Factoring x² - 4x - 5, we get:
(x - 5)(x + 1) > 0
Now, we find the critical points by setting each factor equal to zero:
x - 5 = 0 => x = 5
x + 1 = 0 => x = -1
So, our critical points are x = 5 and x = -1. These points divide the number line into three intervals: (-∞, -1), (-1, 5), and (5, ∞). Now we need to test a value from each interval to see if it satisfies the original inequality. Choose a test value in each interval:
- For (-∞, -1), let's pick x = -2:
(-2 - 5)(-2 + 1) = (-7)(-1) = 7 > 0 (This interval satisfies the inequality) 2. For (-1, 5), let's pick x = 0:
(0 - 5)(0 + 1) = (-5)(1) = -5 < 0 (This interval does not satisfy the inequality) 3. For (5, ∞), let's pick x = 6:
(6 - 5)(6 + 1) = (1)(7) = 7 > 0 (This interval satisfies the inequality)
Since we want the intervals where x² - 4x - 5 > 0, we include the intervals (-∞, -1) and (5, ∞). We use parentheses because the inequality is strictly greater than zero, meaning we don't include the endpoints. So, the solution in interval notation is:
(-∞, -1) ∪ (5, ∞)
The ∪ symbol means "union," which means we're combining the two intervals. Expressing solutions in interval notation is a concise way to show the range of values that satisfy an inequality. Remember to factor, find critical points, test intervals, and use the correct notation (parentheses or brackets) to indicate whether endpoints are included. With a little practice, you'll get the hang of it! Understanding how to represent solutions using intervals is super useful, especially when you're dealing with more advanced math topics. It might seem a bit tricky at first, but trust me, it’s a skill worth mastering.
Now, let’s move on to simplifying rational expressions. Rational expressions are just fractions where the numerator and denominator are polynomials. Simplifying them is all about canceling out common factors, just like you do with regular fractions. This skill is super important in algebra and calculus. When you simplify rational expressions, you're making them easier to work with. This can save you a lot of time and effort in more complex problems. Plus, it’s kind of like tidying up your math work – making everything look cleaner and more manageable. So, let's dive in and see how it's done.
Example 3: Simplifying a Basic Rational Expression
We don't have a specific expression for this example but let's use 2x/(x+1) + (x-5)/(x² - x - 20) in the next example. However, to simplify a basic rational expression, you first factor both the numerator and the denominator as much as possible. Then, you cancel out any factors that appear in both. For instance, if you have (x+2)(x-1) / (x+2)(x+3), you can cancel out the (x+2) factors, leaving you with (x-1) / (x+3). Remember, you can only cancel factors, not terms. A factor is something that’s multiplied, while a term is something that’s added or subtracted. This is a common mistake, so watch out for it!
Example 3.1: 2x/(x+1) + (x-5)/(x² - x - 20)
Let's tackle a more complex example: 2x/(x+1) + (x-5)/(x² - x - 20). This problem involves adding two rational expressions, which means we need a common denominator first. The first fraction already has a denominator of (x+1). To find the common denominator, we need to factor the denominator of the second fraction:
x² - x - 20
Factoring this quadratic expression, we get:
(x - 5)(x + 4)
So, the second fraction is (x-5) / ((x - 5)(x + 4)). Now we can see that the common denominator for both fractions will be (x + 1)(x - 5)(x + 4). We need to rewrite each fraction with this common denominator. For the first fraction, 2x/(x+1), we need to multiply both the numerator and the denominator by (x - 5)(x + 4):
[2x(x - 5)(x + 4)] / [(x + 1)(x - 5)(x + 4)]
For the second fraction, (x-5) / ((x - 5)(x + 4)), we need to multiply both the numerator and the denominator by (x + 1):
[(x - 5)(x + 1)] / [(x - 5)(x + 4)(x + 1)]
Now we have:
[2x(x - 5)(x + 4)] / [(x + 1)(x - 5)(x + 4)] + [(x - 5)(x + 1)] / [(x - 5)(x + 4)(x + 1)]
Next, we combine the numerators over the common denominator:
[2x(x - 5)(x + 4) + (x - 5)(x + 1)] / [(x + 1)(x - 5)(x + 4)]
Now, we need to expand and simplify the numerator. This part can get a bit messy, so let’s take it step by step. First, expand 2x(x - 5)(x + 4):
2x(x² - x - 20) = 2x³ - 2x² - 40x
Then, expand (x - 5)(x + 1):
x² - 4x - 5
Now, combine these results:
2x³ - 2x² - 40x + x² - 4x - 5 = 2x³ - x² - 44x - 5
So, our expression becomes:
(2x³ - x² - 44x - 5) / [(x + 1)(x - 5)(x + 4)]
Finally, we check if we can factor the numerator to see if there are any common factors with the denominator. In this case, the numerator 2x³ - x² - 44x - 5 does not factor nicely, so we can’t simplify further. Thus, the simplified expression is:
(2x³ - x² - 44x - 5) / [(x + 1)(x - 5)(x + 4)]
Simplifying rational expressions can be a bit involved, but the key is to take it one step at a time. Factor the denominators, find a common denominator, combine the fractions, simplify the numerator, and then see if you can cancel any common factors. It’s like solving a puzzle – each step brings you closer to the final solution. Practice makes perfect, so keep at it!
So, guys, we've covered a lot in this article! We tackled solving quadratic equations using the quadratic formula, expressing solutions for quadratic inequalities in interval notation, and simplifying rational expressions. These are fundamental skills in algebra, and mastering them will set you up for success in more advanced math courses. Remember, math is like building blocks – each concept builds on the previous one. The quadratic formula is your go-to for solving equations that don't factor easily. Interval notation is a neat way to express the solutions of inequalities. And simplifying rational expressions is all about making complex fractions more manageable. Don't get discouraged if it seems tough at first. Math takes practice, and everyone learns at their own pace. Keep reviewing the steps, work through examples, and don't be afraid to ask for help when you need it. With dedication and effort, you'll get there! And hey, math can actually be pretty fun once you start to see how everything fits together. So, keep exploring, keep learning, and most importantly, keep believing in yourself.