Solve $x^2+8x=33$ By Completing The Square

by Viktoria Ivanova 43 views

Hey guys! Let's dive into solving quadratic equations, specifically using the completing the square method. This technique is super useful, especially when you can't easily factor the equation. We'll break down the process step by step, making it crystal clear how to tackle these problems. We'll focus on the equation x2+8x=33x^2 + 8x = 33 and figure out its solution set. So, buckle up and let’s get started!

Understanding Completing the Square

Before we jump into the equation x2+8x=33x^2 + 8x = 33, let's quickly recap what completing the square actually means. At its core, completing the square is a method used to rewrite a quadratic equation in a form that allows us to easily find the solutions. The goal is to transform the quadratic expression into a perfect square trinomial, which can then be factored into a binomial squared. This makes solving for the variable much simpler.

Think of it like this: We want to manipulate our equation so that one side looks like (x+a)2(x + a)^2 or (xβˆ’a)2(x - a)^2. When we expand these, we get x2+2ax+a2x^2 + 2ax + a^2 and x2βˆ’2ax+a2x^2 - 2ax + a^2 respectively. By completing the square, we're essentially figuring out what that 'a' value should be and adding the necessary term to both sides of the equation to create this perfect square trinomial. This transformation is incredibly powerful because once we have a perfect square, we can simply take the square root of both sides and solve for x.

Now, you might be wondering why this method is so important. Well, completing the square not only helps us solve quadratic equations, but it's also a key concept in deriving the quadratic formula itself. Plus, it's a fundamental technique in various areas of mathematics, including calculus and conic sections. Mastering completing the square gives you a solid foundation for more advanced topics, so it's well worth the effort to understand it thoroughly. It provides a structured approach to solving equations that might otherwise seem daunting, and it enhances your overall problem-solving skills in mathematics.

Step-by-Step Solution for x2+8x=33x^2 + 8x = 33

Okay, let's apply this method to our equation: x2+8x=33x^2 + 8x = 33. Follow these steps closely, and you'll see how smoothly it works!

Step 1: Prepare the Equation

The first thing we need to do is make sure our equation is in the standard form for completing the square. This usually means having the quadratic and linear terms on one side and the constant term on the other. In our case, the equation x2+8x=33x^2 + 8x = 33 is already in this form. We have the x2x^2 term and the 8x8x term on the left side, and the constant 33 on the right side. So, we can move straight to the next step without any rearranging. This initial setup is crucial because it sets the stage for the subsequent steps, ensuring that we can correctly identify the necessary components for completing the square.

Step 2: Find the Value to Complete the Square

This is where the magic happens! We need to find a value that, when added to our left side, will turn it into a perfect square trinomial. The key here is to focus on the coefficient of the x term. In our equation, the coefficient of x is 8. To find the value we need to add, we take half of this coefficient and then square it. So, we calculate (8/2)2=42=16(8 / 2)^2 = 4^2 = 16. This means that 16 is the value we need to add to both sides of the equation to complete the square. Remember, this step is based on the formula (b/2)2(b/2)^2, where b is the coefficient of the x term. Understanding this formula makes the process much clearer and easier to remember. This value will allow us to rewrite the left side as a perfect square, which is the whole point of this method.

Step 3: Add the Value to Both Sides

Now that we've found the value needed to complete the square, we need to add it to both sides of the equation. This is crucial to maintain the balance of the equationβ€”what we do to one side, we must also do to the other. So, we add 16 to both sides of x2+8x=33x^2 + 8x = 33, which gives us x2+8x+16=33+16x^2 + 8x + 16 = 33 + 16. This step ensures that the equation remains equivalent to its original form, while also moving us closer to our goal of creating a perfect square trinomial on the left side. By adding the same value to both sides, we preserve the equality and set up the equation for the next crucial step: factoring.

Step 4: Factor the Left Side

The left side of our equation, x2+8x+16x^2 + 8x + 16, is now a perfect square trinomial. This means it can be factored into the form (x+a)2(x + a)^2. In our case, x2+8x+16x^2 + 8x + 16 factors into (x+4)2(x + 4)^2. To verify this, you can expand (x+4)2(x + 4)^2 and see that it indeed equals x2+8x+16x^2 + 8x + 16. Factoring the perfect square trinomial is a key step in solving the quadratic equation because it simplifies the equation into a form where we can easily isolate x. This step transforms the quadratic equation into a more manageable algebraic expression, making it straightforward to find the solutions.

Step 5: Simplify the Right Side

While we were busy factoring the left side, we also added 16 to the right side of the equation. So, we need to simplify 33+1633 + 16, which equals 49. Our equation now looks like (x+4)2=49(x + 4)^2 = 49. Simplifying the right side makes the equation cleaner and prepares it for the next operation, which is taking the square root of both sides. This simplification ensures that we have a clear and manageable equation to work with as we proceed towards finding the values of x. This step is crucial for setting up the final stages of the solution.

Step 6: Take the Square Root of Both Sides

To get rid of the square on the left side, we take the square root of both sides of the equation. Remember, when we take the square root, we need to consider both the positive and negative roots. So, the square root of (x+4)2(x + 4)^2 is x+4x + 4, and the square root of 49 is both +7 and -7. This gives us two equations: x+4=7x + 4 = 7 and x+4=βˆ’7x + 4 = -7. Remembering to consider both positive and negative roots is crucial because quadratic equations often have two solutions. This step is pivotal in uncovering both possible values for x and completing the solution process.

Step 7: Solve for x

Now we have two simple equations to solve: x+4=7x + 4 = 7 and x+4=βˆ’7x + 4 = -7. For the first equation, we subtract 4 from both sides to get x=7βˆ’4x = 7 - 4, which simplifies to x=3x = 3. For the second equation, we subtract 4 from both sides to get x=βˆ’7βˆ’4x = -7 - 4, which simplifies to x=βˆ’11x = -11. So, we have found two solutions for x: 3 and -11. These values are the roots of the quadratic equation and represent the points where the parabola intersects the x-axis. Solving for x in these two simple equations is the final step in finding the solution set for the original quadratic equation.

Solution Set

The solutions we found are x=3x = 3 and x=βˆ’11x = -11. Therefore, the solution set for the equation x2+8x=33x^2 + 8x = 33 is {βˆ’11,3}\{-11, 3\}. This means that both -11 and 3 satisfy the original equation. You can verify this by plugging each value back into the original equation and confirming that it holds true. The solution set represents all the values of x that make the equation a true statement. In this case, the solution set {βˆ’11,3}\{-11, 3\} provides a concise way to express the two roots of the quadratic equation, completing our step-by-step solution.

Why Completing the Square Matters

You might be wondering,