Solving Quadratic Equations: Find The Solutions For (x+3)^2 = 49

Hey guys! Let's dive into solving a quadratic equation. Today, we're tackling the equation $(x+3)^2 = 49$. Quadratic equations might seem intimidating at first, but with a step-by-step approach, they become much easier to handle. So, let's break it down and find the solutions together.

Understanding Quadratic Equations

First off, what exactly is a quadratic equation? Quadratic equations are polynomial equations of the second degree, meaning the highest power of the variable (in our case, x) is 2. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where a, b, and c are constants, and a is not equal to zero. These equations pop up in various areas of mathematics and physics, so understanding how to solve them is super important. Before we jump into our specific problem, it's good to know that there are a few common methods for solving quadratic equations:

• Factoring: This involves breaking down the quadratic expression into two binomials. If you can factor the equation, you can easily find the solutions by setting each factor equal to zero.
• Completing the Square: This method involves manipulating the equation to form a perfect square trinomial, which can then be easily solved.
• Quadratic Formula: This is a general formula that works for any quadratic equation. It's especially useful when factoring or completing the square is tricky.

Now that we have a basic understanding, let's get back to our equation: $(x+3)^2 = 49$.

Solving $(x+3)^2 = 49$: Step-by-Step

Method 1: Taking the Square Root

The equation $(x+3)^2 = 49$ is already in a nice form for us to use the square root method. This method is super handy when you have a perfect square on one side of the equation. Here’s how it works:

1. Take the square root of both sides:

   When you take the square root of a squared term, you get the original term back. But remember, when you take the square root of a number, you need to consider both the positive and negative roots. So, we have:

   $\\\sqrt{(x+3)^2} = \\\ ext{±}\\\\sqrt{49}$

   This simplifies to:

   $x + 3 = \\\ ext{±}7$

2. Separate into two equations:

   Now we have two possible equations:

   • $x + 3 = 7$
   • $x + 3 = -7$

3. Solve each equation for x:

   For the first equation, $x + 3 = 7$, subtract 3 from both sides:

   $x = 7 - 3$

   $x = 4$

   For the second equation, $x + 3 = -7$, subtract 3 from both sides:

   $x = -7 - 3$

   $x = -10$

So, the solutions are $x = 4$ and $x = -10$.

Method 2: Expanding and Factoring

Another way to solve this equation is by expanding the square and then either factoring or using the quadratic formula. Let’s walk through that:

1. Expand the left side:

   $(x+3)^2$ means $(x+3)(x+3)$. Multiply this out:

   $(x+3)(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9$

   So, our equation becomes:

   $x^2 + 6x + 9 = 49$

2. Move all terms to one side:

   Subtract 49 from both sides to set the equation to zero:

   $x^2 + 6x + 9 - 49 = 0$

   $x^2 + 6x - 40 = 0$

3. Factor the quadratic expression:

   We’re looking for two numbers that multiply to -40 and add to 6. Those numbers are 10 and -4. So, we can factor the expression as:

   $(x + 10)(x - 4) = 0$

4. Set each factor equal to zero:

   • $x + 10 = 0$
   • $x - 4 = 0$

5. Solve for x:

   For $x + 10 = 0$, subtract 10 from both sides:

   $x = -10$

   For $x - 4 = 0$, add 4 to both sides:

   $x = 4$

Again, we find the solutions are $x = 4$ and $x = -10$.

Method 3: Using the Quadratic Formula

If factoring isn't straightforward, the quadratic formula is your best friend. Remember the quadratic formula? It’s:

$x = \\frac{-b \\\ ext{±} \\\\sqrt{b^2 - 4ac}}{2a}$

For our equation $x^2 + 6x - 40 = 0$, we have:

• $a = 1$
• $b = 6$
• $c = -40$

Plug these values into the quadratic formula:

$x = \\frac{-6 \\\ ext{±} \\\\sqrt{6^2 - 4(1)(-40)}}{2(1)}$

$x = \\frac{-6 \\\ ext{±} \\\\sqrt{36 + 160}}{2}$

$x = \\frac{-6 \\\ ext{±} \\\\sqrt{196}}{2}$

$x = \\frac{-6 \\\ ext{±} 14}{2}$

Now, we have two possibilities:

1. Using the + sign:

   $x = \\frac{-6 + 14}{2} = \\frac{8}{2} = 4$

2. Using the - sign:

   $x = \\frac{-6 - 14}{2} = \\frac{-20}{2} = -10$

Once again, the solutions are $x = 4$ and $x = -10$.

Verifying the Solutions

It's always a good idea to check your answers to make sure they're correct. Let's plug our solutions back into the original equation $(x+3)^2 = 49$:

For $x = 4$:

$(4+3)^2 = 7^2 = 49$

So, $x = 4$ is indeed a solution.

For $x = -10$:

$(-10+3)^2 = (-7)^2 = 49$

And $x = -10$ is also a solution.

Conclusion

So, there you have it! The solutions to the quadratic equation $(x+3)^2 = 49$ are $x = 4$ and $x = -10$. We solved it using three different methods: taking the square root, expanding and factoring, and using the quadratic formula. Each method gives us the same answers, which is always reassuring. Remember, the key to mastering quadratic equations is practice, so keep at it, and you'll become a pro in no time! If you have more questions or want to explore other types of equations, just keep exploring and learning. You got this!