Solve Quadratic Equations Easily

by Natalie Brooks 33 views

Hey guys! Let's dive into the exciting world of equation solving. We're going to tackle a set of equations, breaking them down step-by-step so you can see exactly how it's done. Whether you're a student brushing up on your algebra skills or just someone who enjoys a good mathematical challenge, this guide is for you. We'll be working through equations a, b, c, and d, ensuring you grasp the techniques involved. So, grab your pencils and let's get started!

a. Solving the Quadratic Equation: 49x² + 28x + 4 = 0

Okay, let's kick things off with the first equation: 49x² + 28x + 4 = 0. This looks like a quadratic equation, and the key to solving these is often recognizing patterns or using the quadratic formula. But before we jump into the formula, let’s see if we can simplify this a bit. Notice anything special about the coefficients? They might just be hinting at something cool!

  • Spotting the Perfect Square Trinomial: If you look closely, you'll see that 49x² is (7x)², 4 is 2², and 28x is 2 * (7x) * 2. This means we've got ourselves a perfect square trinomial! These are awesome because they can be factored super easily. A perfect square trinomial follows the pattern a² + 2ab + b² = (a + b)². In our case, a is 7x and b is 2. So, we can rewrite the equation as (7x + 2)² = 0.

  • Taking the Square Root: Now that we’ve got it in this neat form, the next step is straightforward. We take the square root of both sides. The square root of (7x + 2)² is simply 7x + 2, and the square root of 0 is, well, still 0. This gives us the equation 7x + 2 = 0. See how much simpler that looks?

  • Isolating x: Almost there! To solve for x, we need to isolate it. First, we subtract 2 from both sides of the equation, giving us 7x = -2. Then, we divide both sides by 7, which finally gives us x = -2/7. And that's our solution! We've successfully solved the first quadratic equation by recognizing the perfect square trinomial pattern.

    So, the solution to the equation 49x² + 28x + 4 = 0 is x = -2/7. Wasn't that satisfying? Let's move on to the next one!

b. Tackling Another Quadratic: 5x² - 54 = 0

Alright, let's move on to the second equation: 5x² - 54 = 0. This one looks a little different from the first one. We don’t have the 'x' term in the middle, which actually makes it a bit simpler to solve. We can isolate the x² term directly.

  • Isolating x²: Our first goal is to get the x² term by itself on one side of the equation. To do this, we add 54 to both sides, which gives us 5x² = 54. Now, we're one step closer to solving for x.

  • Dividing to Simplify: Next, we want to get x² completely alone, so we divide both sides by 5. This gives us x² = 54/5. We're almost there! We've got x² isolated, and now we just need to undo that square.

  • Taking the Square Root (Don't Forget the ±): This is a crucial step! To get x by itself, we take the square root of both sides. Remember, whenever we take the square root in solving an equation, we need to consider both the positive and negative roots. So, we have x = ±√(54/5). We're not quite done yet, though. Let's simplify that radical.

  • Simplifying the Radical: The square root of a fraction can be a bit messy, so let's break it down. √(54/5) can be written as √54 / √5. We can simplify √54 by factoring out a perfect square. 54 is 9 * 6, and 9 is a perfect square (3²). So, √54 = √(9 * 6) = 3√6. Now we have x = ±(3√6 / √5). To get rid of the square root in the denominator, we'll rationalize it by multiplying both the numerator and denominator by √5. This gives us x = ±(3√6 * √5) / (√5 * √5) = ±(3√30) / 5.

    So, the solutions to the equation 5x² - 54 = 0 are x = (3√30) / 5 and x = -(3√30) / 5. We’ve tackled another one! Let’s move on to the next equation.

c. Solving by Factoring: 6x² + 18x = 0

Now, let’s tackle the equation 6x² + 18x = 0. This one looks a bit different again, and that's great! It gives us a chance to use a different technique: factoring. Factoring is a powerful tool for solving equations, especially when we see common factors.

  • Identifying the Common Factor: When we look at 6x² and 18x, what do we see in common? Well, both terms are divisible by 6, and they both have an x. So, the common factor is 6x. Factoring out 6x from both terms is our first step.

  • Factoring Out the Common Factor: We pull out 6x from the equation: 6x(x + 3) = 0. See how we've rewritten the equation as a product of two factors? This is the key to solving it. When the product of two factors is zero, at least one of the factors must be zero. It’s like saying, “If A times B is zero, then either A is zero, or B is zero, or both!”

  • Setting Each Factor to Zero: Now we set each factor equal to zero and solve for x. So, we have two equations: 6x = 0 and x + 3 = 0.

  • Solving for x: For the first equation, 6x = 0, we divide both sides by 6, which gives us x = 0. For the second equation, x + 3 = 0, we subtract 3 from both sides, which gives us x = -3. And there we have it – two solutions!

    So, the solutions to the equation 6x² + 18x = 0 are x = 0 and x = -3. Factoring saved the day! Let’s move on to our last equation.

d. Expanding and Simplifying: (x - 3)(2x - 1) + (x + 2)(2x + 1) = 17

Last but not least, we have the equation (x - 3)(2x - 1) + (x + 2)(2x + 1) = 17. This one looks a bit more involved, but don't worry, we'll break it down. The key here is to first expand the products and then simplify the equation. Let's get started!

  • Expanding the Products: We need to multiply out those binomials. Remember the FOIL method (First, Outer, Inner, Last)? That’s our friend here. Let's expand (x - 3)(2x - 1) first. First: x * 2x = 2x². Outer: x * -1 = -x. Inner: -3 * 2x = -6x. Last: -3 * -1 = 3. So, (x - 3)(2x - 1) = 2x² - x - 6x + 3 = 2x² - 7x + 3.

    Now, let's expand (x + 2)(2x + 1). First: x * 2x = 2x². Outer: x * 1 = x. Inner: 2 * 2x = 4x. Last: 2 * 1 = 2. So, (x + 2)(2x + 1) = 2x² + x + 4x + 2 = 2x² + 5x + 2.

  • Combining Like Terms: Now we substitute these expanded forms back into the original equation: (2x² - 7x + 3) + (2x² + 5x + 2) = 17. Let's combine like terms. We have 2x² + 2x² = 4x², -7x + 5x = -2x, and 3 + 2 = 5. So, our equation simplifies to 4x² - 2x + 5 = 17.

  • Setting the Equation to Zero: To solve this quadratic equation, we need to set it equal to zero. Subtract 17 from both sides: 4x² - 2x + 5 - 17 = 0, which simplifies to 4x² - 2x - 12 = 0. We're getting there! Now, let’s see if we can simplify further by dividing out a common factor.

  • Simplifying the Equation: Notice that all the coefficients are even? We can divide the entire equation by 2 to make things a bit easier. This gives us 2x² - x - 6 = 0.

  • Factoring or Using the Quadratic Formula: Now we have a simpler quadratic equation. We can try to factor it, or if that’s tricky, we can use the quadratic formula. Let’s try factoring first. We’re looking for two numbers that multiply to -12 (2 * -6) and add up to -1. After a bit of thought, we can see that -4 and 3 fit the bill. So, we can rewrite the middle term as -4x + 3x: 2x² - 4x + 3x - 6 = 0.

  • Factoring by Grouping: Now we factor by grouping. From the first two terms, we can factor out 2x: 2x(x - 2). From the last two terms, we can factor out 3: 3(x - 2). So, we have 2x(x - 2) + 3(x - 2) = 0. Now we can factor out the common factor (x - 2): (x - 2)(2x + 3) = 0.

  • Setting Each Factor to Zero and Solving: Now we set each factor to zero and solve: x - 2 = 0 and 2x + 3 = 0. For x - 2 = 0, we add 2 to both sides, giving us x = 2. For 2x + 3 = 0, we subtract 3 from both sides, giving us 2x = -3, and then divide by 2, giving us x = -3/2.

    So, the solutions to the equation (x - 3)(2x - 1) + (x + 2)(2x + 1) = 17 are x = 2 and x = -3/2. Whew! We made it through that one!

Conclusion: You've Got This!

And there you have it, guys! We've solved four different equations using a variety of techniques: recognizing perfect square trinomials, isolating variables, factoring, and expanding and simplifying. Each equation gave us a chance to flex our mathematical muscles, and hopefully, you've picked up some valuable skills along the way.

Solving equations is a fundamental part of algebra, and the more you practice, the better you'll get. So, keep at it, and don't be afraid to tackle even the trickiest-looking problems. With a little patience and the right techniques, you can conquer any equation that comes your way. Happy solving!