How To Solve X² + 15x = -56: A Step-by-Step Guide

When faced with a quadratic equation like $x^2+15 x=-56$, it's natural to wonder how to find the values of $x$ that make the equation true. This type of problem is a staple in algebra, and understanding how to solve it opens the door to tackling more complex mathematical challenges. We're going to break down this specific equation, $x^2+15 x=-56$, and walk you through the process of finding its solutions. Our goal is to make this seem less like a daunting task and more like a puzzle you can solve with a few key algebraic steps. So, let's dive in and uncover the secrets hidden within this quadratic expression. We'll cover rearranging the equation into its standard form, factoring, and finally, determining the correct solutions for $x$. By the end, you'll see that solving $x^2+15 x=-56$ is well within your reach, and you'll gain a valuable skill applicable to many other mathematical scenarios. Ready to get started? Let's tackle this equation head-on and find those elusive values of $x$ that satisfy the condition.

Understanding Quadratic Equations and Standard Form

Before we can solve $x^2+15 x=-56$, it's crucial to understand what a quadratic equation is and why we aim to put it into a specific format. A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one term that is squared. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and importantly, $a$ cannot be zero. This standard form, $ax^2 + bx + c = 0$, is essential because it provides a consistent structure that allows us to apply various solving methods, such as factoring or using the quadratic formula. Our equation, $x^2+15 x=-56$, isn't quite in this standard form yet because the constant term (-56) is on the right side of the equals sign. To transform it, we need to move all terms to one side, leaving zero on the other. This is a fundamental step in solving most quadratic equations, as it sets the stage for manipulation and calculation. By adding 56 to both sides of the equation, we effectively isolate the zero on the right, giving us $x^2+15 x+56=0$. Now, this equation is in the standard quadratic form, where $a=1$, $b=15$, and $c=56$. This structured approach ensures that we are working with a consistent framework, making it easier to choose and apply the most efficient method for finding the roots, or solutions, of the equation. Mastering the ability to convert equations into this standard form is a cornerstone of algebraic problem-solving, paving the way for more advanced concepts and applications.

Factoring the Quadratic Expression

With our equation rearranged to the standard form, $x^2+15 x+56=0$, the next logical step is to attempt to factor the quadratic expression. Factoring is a method of rewriting a polynomial as a product of simpler expressions, typically binomials. For a quadratic of the form $x^2+bx+c$, we're looking for two numbers that multiply to equal $c$ and add up to equal $b$. In our case, we need two numbers that multiply to 56 and add up to 15. This process involves a bit of trial and error, or a systematic approach. Let's list the pairs of factors for 56: (1, 56), (2, 28), (4, 14), and (7, 8). Now, let's check the sum of each pair: 1 + 56 = 57, 2 + 28 = 30, 4 + 14 = 18, and 7 + 8 = 15. Eureka! The pair (7, 8) fits both conditions: $7 imes 8 = 56$ and $7 + 8 = 15$. Since we found these two numbers, we can now rewrite our quadratic equation in its factored form. The factored form will be $(x+7)(x+8)=0$. This is a much simpler expression to work with. The principle behind factoring is that if the product of two (or more) factors is zero, then at least one of those factors must be zero. This concept is known as the Zero Product Property, and it's the key to unlocking the solutions for $x$. By setting each factor equal to zero, we can isolate $x$ and find the values that satisfy the original equation. This method is often the quickest way to solve quadratic equations, provided the expression can be factored neatly. It's a testament to the elegance of algebraic manipulation that we can transform a complex-looking equation into a product of simple terms, thereby revealing its solutions.

Finding the Solutions for x

Now that we have successfully factored the quadratic equation $x^2+15 x+56=0$ into $(x+7)(x+8)=0$, we can use the Zero Product Property to find the solutions for $x$. This property states that if the product of two or more factors is zero, then at least one of the factors must be zero. In our case, we have two factors: $(x+7)$ and $(x+8)$. For their product to be zero, either $(x+7)$ must equal zero, or $(x+8)$ must equal zero (or both, but that's not possible here as $x$ can only have one value at a time within each factor). So, we set up two separate, simple linear equations:

1. $x+7 = 0$
2. $x+8 = 0$

Solving the first equation, $x+7=0$, we subtract 7 from both sides to isolate $x$: $x = -7$. This is one solution.

Solving the second equation, $x+8=0$, we subtract 8 from both sides to isolate $x$: $x = -8$. This is our second solution.

Therefore, the solutions to the equation $x^2+15 x=-56$ are $x=-7$ and $x=-8$. These are the values of $x$ that, when substituted back into the original equation, will make the statement true. To verify these solutions, you can plug them back into the original equation:

For $x=-7$: $(-7)^2 + 15(-7) = 49 - 105 = -56$. This is correct.

For $x=-8$: $(-8)^2 + 15(-8) = 64 - 120 = -56$. This is also correct.

Seeing these solutions confirmed reinforces the power of algebraic methods. We've moved from a seemingly complex equation to two concrete answers through logical steps of rearrangement, factoring, and applying fundamental properties. This methodical approach is what makes mathematics so powerful and predictable. The specific options provided in the original question often include correct solutions alongside distractors, so it's essential to perform the calculation correctly to select the right answer. In this case, the correct pair of solutions is $x=-8$ and $x=-7$. This methodical process ensures accuracy and builds confidence in your mathematical abilities.

Considering Alternative Methods: The Quadratic Formula

While factoring is often the most straightforward method for solving quadratic equations like $x^2+15 x=-56$, it's not always possible to factor easily, or at all, using integers. Fortunately, there's a universal tool that can solve any quadratic equation: the quadratic formula. This formula is derived from the process of completing the square on the general quadratic equation $ax^2 + bx + c = 0$. The formula is given by: x = rac{-b oon sqrt{b^2-4ac}}{2a}. Even though we successfully factored our equation, let's use the quadratic formula to demonstrate its application and to confirm our solutions. Recall that our equation in standard form is $x^2+15 x+56=0$, which means $a=1$, $b=15$, and $c=56$. Plugging these values into the quadratic formula, we get:

x = rac{-15 oon sqrt{15^2-4(1)(56)}}{2(1)}

Now, let's simplify the expression under the square root (the discriminant):

$15^2 = 225$

$4(1)(56) = 4 imes 56 = 224$

So, the discriminant is $b^2-4ac = 225 - 224 = 1$.

Substituting this back into the formula:

x = rac{-15 oon sqrt{1}}{2}

x = rac{-15 oon 1}{2}

This gives us two possible solutions, one for the plus sign and one for the minus sign:

1. Using the plus sign: x = rac{-15 + 1}{2} = rac{-14}{2} = -7
2. Using the minus sign: x = rac{-15 - 1}{2} = rac{-16}{2} = -8

As you can see, the quadratic formula yields the exact same solutions: $x=-7$ and $x=-8$. This confirms that our factoring method was correct and also highlights the power and reliability of the quadratic formula. It's a crucial tool in any mathematician's arsenal, especially when dealing with equations that don't factor nicely or when you need a guaranteed method to find the roots. Understanding both factoring and the quadratic formula provides a comprehensive approach to solving quadratic equations, equipping you with flexibility and confidence.

Conclusion: Finding the Right Solutions

We've successfully navigated the process of solving the quadratic equation $x^2+15 x=-56$. By first rearranging the equation into its standard form, $x^2+15 x+56=0$, we prepared it for common solving techniques. We then employed the powerful method of factoring, identifying that two numbers, 7 and 8, multiply to 56 and add to 15. This allowed us to rewrite the equation as $(x+7)(x+8)=0$. Applying the Zero Product Property, we set each factor to zero, yielding the solutions $x=-7$ and $x=-8$. We also confirmed these results using the universal quadratic formula, demonstrating its efficacy and providing an alternative pathway to the same correct answers. This methodical approach is key to mastering algebraic problems. When presented with options like A. $x=8$ and $x=7$, B. $x=-8$ and $x=-7$, C. $x=-8$ and $x=7$, or D. $x=8$ and $x=-7$, our calculated solutions clearly point to option B. Understanding how to solve quadratic equations is a fundamental skill in mathematics, applicable to numerous real-world scenarios, from calculating trajectories to optimizing financial models. Keep practicing these methods, and you'll find that solving equations becomes increasingly intuitive and rewarding.

For further exploration into the fascinating world of algebra and quadratic equations, you can visit Khan Academy's comprehensive resources on algebra, or delve into the detailed explanations on MathWorld. These sites offer a wealth of information, practice problems, and tutorials that can deepen your understanding and hone your skills.