Solving Y² + 4y + 4 = 7 A Step By Step Guide

Hey guys! Today, we're diving into a classic math problem: solving a quadratic equation. Specifically, we're going to find the value or values of y in the equation y² + 4y + 4 = 7. Quadratic equations might seem intimidating at first, but with the right approach, they can be quite straightforward. We'll explore different methods to solve this, ensuring you understand each step along the way. Whether you're a student brushing up on algebra or just a curious mind, this guide will help you master this type of problem. So, let's get started and break down this equation together!

Before we jump into solving the equation y² + 4y + 4 = 7, let's take a moment to understand what quadratic equations are all about. A quadratic equation is essentially a polynomial equation of the second degree. This means the highest power of the variable (in our case, y) is 2. The general form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants, and a is not equal to zero. If a were zero, the equation would become linear, not quadratic.

Now, you might be wondering, why are these equations so important? Well, quadratic equations pop up in various fields, from physics and engineering to economics and computer science. They can describe the trajectory of a projectile, the shape of a satellite dish, or even model growth and decay processes. Understanding how to solve them is a fundamental skill in mathematics and a valuable tool in many practical applications.

Our equation, y² + 4y + 4 = 7, fits the quadratic form, though it's not immediately in the standard form of ax² + bx + c = 0. The first step in solving it will be to rearrange it into this standard form. We'll see how to do that in the next section. By recognizing the structure of a quadratic equation, we can apply various techniques to find the values of y that satisfy the equation. These values are often called the roots or solutions of the equation.

There are several methods we can use to solve quadratic equations, each with its own strengths and when it's most applicable. Let's briefly touch on some of the most common methods before we apply one to our equation, y² + 4y + 4 = 7. This overview will help you build a solid foundation for tackling any quadratic equation that comes your way.

1. Factoring: This method involves expressing the quadratic expression as a product of two binomials. For example, if we can rewrite ax² + bx + c as (px + q) (rx + s), we can then set each factor equal to zero and solve for x. Factoring is often the quickest method when the quadratic expression is easily factorable.

2. Completing the Square: This method involves manipulating the equation to form a perfect square trinomial on one side. It's a powerful technique that can be used to solve any quadratic equation, even those that are difficult to factor. Completing the square is especially useful when the coefficient of the x² term (a) is 1.

3. Quadratic Formula: This is a universal formula that provides the solutions for any quadratic equation. The formula is x = [-b ± √(b² - 4ac)] / (2a), where a, b, and c are the coefficients from the standard form of the equation. The quadratic formula is a reliable method when factoring is not straightforward or when you simply want a direct approach.

4. Graphical Method: While not as precise as the algebraic methods, graphing the quadratic equation (which forms a parabola) can visually show the solutions as the points where the parabola intersects the x-axis. This method is great for understanding the nature of the solutions but might not give exact values.

For our equation, y² + 4y + 4 = 7, we'll explore a method that combines both factoring principles and a bit of algebraic manipulation to arrive at the solutions. Knowing these different methods gives you a flexible toolkit for solving quadratic equations.

Alright, let's dive into solving our specific equation: y² + 4y + 4 = 7. The first step, as we mentioned earlier, is to get the equation into the standard quadratic form, which is ay² + by + c = 0. Currently, we have 7 on the right side, so we need to move it to the left side of the equation.

To do this, we subtract 7 from both sides of the equation. This gives us:

y² + 4y + 4 - 7 = 7 - 7

Simplifying this, we get:

y² + 4y - 3 = 0

Now our equation is in the standard quadratic form, where a = 1, b = 4, and c = -3. Next, we'll attempt to solve this by factoring. We're looking for two numbers that multiply to c (-3) and add up to b (4). This might require a bit of trial and error, but it's a crucial skill in algebra.

Think about the factors of -3. They are -1 and 3, or 1 and -3. Which pair adds up to 4? It's 1 and 3 that would add to 4 if we have -1 and +3. Unfortunately, it doesn't quite fit our needs since 1 multiplied by -3 results in -3, but -1 multiplied by 3 also results in -3. However, -1 + 3 would give us 2, not 4. Since we can't find two integers that satisfy both conditions, factoring directly might not be the easiest approach here.

Given the difficulty in directly factoring this equation, we might consider other methods such as completing the square or using the quadratic formula. These methods are more robust and can handle quadratic equations that don't factor neatly. We'll explore the completing the square method in the next section to solve this equation.

Since we encountered a bit of a snag trying to factor the equation y² + 4y - 3 = 0 directly, let's explore another powerful method: completing the square. This technique is super useful for any quadratic equation, especially when factoring isn't straightforward. The idea behind completing the square is to manipulate the equation so that one side becomes a perfect square trinomial, which is a trinomial that can be factored into the square of a binomial.

To begin, we focus on the left side of our equation, y² + 4y - 3. We want to transform the y² + 4y part into a perfect square. Remember that a perfect square trinomial has the form (y + k)² = y² + 2ky + k². Notice that the constant term (k²) is the square of half the coefficient of the y term (2k).

In our equation, the coefficient of the y term is 4. So, half of 4 is 2, and squaring 2 gives us 4. This means we want to create a perfect square trinomial of the form y² + 4y + 4. Lucky for us, the equation originally had y² + 4y + 4 on the left side before we subtracted 7! This gives us a hint that we were close to a perfect square from the start.

Let’s rewrite our equation y² + 4y - 3 = 0 a little differently. We can rewrite it as:

(y² + 4y + 4) - 4 - 3 = 0

Notice that we've added and subtracted 4. This doesn't change the value of the equation, but it allows us to group the perfect square trinomial:

(y² + 4y + 4) - 7 = 0

Now we can rewrite the perfect square trinomial as a squared binomial:

(y + 2)² - 7 = 0

Next, we isolate the squared binomial by adding 7 to both sides:

(y + 2)² = 7

Now, we can take the square root of both sides. Remember, when we take the square root, we need to consider both the positive and negative roots:

√((y + 2)²) = ±√7

This simplifies to:

y + 2 = ±√7

Finally, we solve for y by subtracting 2 from both sides:

y = -2 ± √7

So, we have two solutions for y: y = -2 + √7 and y = -2 - √7. These are the values of y that satisfy the original equation. It's pretty cool how completing the square allowed us to find these solutions, right?

Okay, guys, let's wrap things up! We set out to find the value or values of y in the quadratic equation y² + 4y + 4 = 7. We explored the equation, rearranged it into standard form, and then used the completing the square method to find our solutions. It was a bit of a journey, but we got there!

After carefully working through the steps of completing the square, we found that the solutions for y are:

• y = -2 + √7
• y = -2 - √7

These are the two values of y that make the equation y² + 4y + 4 = 7 true. If you plug either of these values back into the original equation, you'll see that both sides are equal. Pretty neat, huh?

So, the correct answer from our options is:

• d) y = -2 + √7, y = -2 - √7

We've successfully navigated a quadratic equation, learned a valuable technique (completing the square), and found our solutions. Quadratic equations might seem tricky at first, but with practice and the right methods, you can totally master them. Keep practicing, keep exploring, and you'll become a quadratic equation whiz in no time! And remember, math can be fun – especially when you crack a tough problem. Keep up the awesome work!