Solve X Using Completing The Square

Hey guys! Today, we're diving deep into a super cool technique in algebra: completing the square. It's a powerful method for solving quadratic equations, and we're going to use it to tackle a specific problem: finding $x$ in the equation $(x+7)(x-9)=25$. Don't worry if this looks a bit intimidating at first; by the end of this article, you'll be a pro at it. We'll break down each step, explain the 'why' behind it, and make sure you're totally comfortable with the process. So, grab your notebooks, and let's get this math party started!

Understanding the Equation and the Goal

First off, let's look at our equation: $(x+7)(x-9)=25$. This is a quadratic equation, meaning it has an $x^2$ term when expanded. Our goal, as mathematicians, is to isolate $x$ and find the value(s) that make this equation true. While there are other ways to solve quadratic equations, like factoring or using the quadratic formula, completing the square offers a unique insight into the structure of these equations. It's particularly useful when factoring is difficult or when you need to rewrite the quadratic in vertex form. The equation is currently not in the standard form $ax^2 + bx + c = 0$, so our first mission is to get it there. This usually involves expanding the brackets and moving all terms to one side. Remember, the beauty of algebra lies in its systematic approach. By following a set of logical steps, we can transform complex-looking problems into manageable ones. So, let's go ahead and expand the left side of our equation. We'll use the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last).

Expanding and Simplifying the Equation

Alright, let's get down to business and expand the expression $(x+7)(x-9)$. This is where we multiply each term in the first bracket by each term in the second bracket.

• First: $x * x = x^2$
• Outer: $x * (-9) = -9x$
• Inner: $7 * x = 7x$
• Last: $7 * (-9) = -63$

So, $(x+7)(x-9)$ expands to $x^2 - 9x + 7x - 63$. Now, we combine the like terms (the terms with $x$): $-9x + 7x = -2x$. This simplifies our expanded expression to $x^2 - 2x - 63$.

Our equation now looks like this: $x^2 - 2x - 63 = 25$. The next step is to get this equation into the standard quadratic form, $ax^2 + bx + c = 0$. To do this, we need to move the 25 from the right side of the equation to the left side. We achieve this by subtracting 25 from both sides:

$x^2 - 2x - 63 - 25 = 25 - 25$

$x^2 - 2x - 88 = 0$

Awesome! We've successfully transformed our original equation into the standard quadratic form. This is a crucial step because the completing the square method is designed to work with equations in this format. We now have a clear $a$, $b$, and $c$ value (where $a=1$, $b=-2$, and $c=-88$), which will be essential for the upcoming steps. Keep in mind that the goal of completing the square is to manipulate the $x^2$ and $x$ terms into a perfect square trinomial, which we can then easily solve. So, while it might seem like extra work to get to this standard form, it sets us up perfectly for the magic of completing the square. Let's move on to the core of the method!

The Art of Completing the Square

Now for the main event: completing the square! Our equation is $x^2 - 2x - 88 = 0$. The idea behind completing the square is to manipulate the $x^2$ and $x$ terms (in our case, $x^2 - 2x$) so that they form a perfect square trinomial. A perfect square trinomial is a quadratic expression that can be factored into the square of a binomial, like $(x+k)^2$ or $(x-k)^2$. Specifically, $(x+k)^2 = x^2 + 2kx + k^2$ and $(x-k)^2 = x^2 - 2kx + k^2$. Notice the relationship between the coefficient of the $x$ term (the '$b$' value) and the constant term (the '$k^2$' value).

To make $x^2 - 2x$ into a perfect square trinomial, we need to add a specific constant term. This constant is found by taking the coefficient of the $x$ term (which is -2), dividing it by 2, and then squaring the result. Let's do that:

• Take the coefficient of $x$: $-2$
• Divide by 2: $-2 / 2 = -1$
• Square the result: $(-1)^2 = 1$

So, the number we need to add is 1. If we add 1 to $x^2 - 2x$, we get $x^2 - 2x + 1$. This expression is a perfect square trinomial because it can be factored as $(x-1)^2$. Isn't that neat?

However, we can't just randomly add 1 to one side of the equation. To keep the equation balanced, whatever we do to one side, we must do to the other. So, we'll add 1 to both sides of our equation $x^2 - 2x - 88 = 0$. But wait, we first need to isolate the $x^2$ and $x$ terms on one side. Let's move the constant term (-88) to the right side first by adding 88 to both sides:

$x^2 - 2x = 88$

Now, we add our special number (1) to both sides:

$x^2 - 2x + 1 = 88 + 1$

$x^2 - 2x + 1 = 89$

On the left side, we now have our perfect square trinomial $x^2 - 2x + 1$, which we know factors into $(x-1)^2$. So, the equation becomes:

$(x-1)^2 = 89$

And there you have it – we've successfully completed the square! This step is arguably the most important, as it transforms the equation into a form where we can easily solve for $x$ using the square root property. This method is so elegant because it converts a quadratic expression that might not be factorable into a neat, squared binomial. The key is always remembering to add the same value to both sides to maintain equality. It's like balancing a scale; whatever you put on one side, you must mirror on the other.

Solving for $x$ using the Square Root Property

We've reached the exciting part where we finally solve for $x$! Our equation is now in the form $(x-1)^2 = 89$. The beauty of this form is that it directly allows us to use the square root property. This property states that if $y^2 = k$, then $y = \pm\sqrt{k}$. In our case, $y$ is $(x-1)$ and $k$ is 89.

So, we take the square root of both sides of the equation:

$\sqrt{(x-1)^2} = \pm\sqrt{89}$

This simplifies to:

$x-1 = \pm\sqrt{89}$

Remember that $\sqrt{89}$ cannot be simplified further into a whole number since 89 is a prime number. Now, to isolate $x$, we simply need to add 1 to both sides of the equation:

$x = 1 \pm\sqrt{89}$

This gives us two possible solutions for $x$:

1. $x = 1 + \sqrt{89}$
2. $x = 1 - \sqrt{89}$

These are the exact solutions for our original equation. You could also calculate approximate decimal values if needed, but for most mathematical purposes, the exact form is preferred. The process of completing the square has allowed us to systematically arrive at these solutions, revealing the two points where the parabola represented by the quadratic equation intersects the x-axis. It's a testament to the power of algebraic manipulation that we can take a seemingly complex equation and distill it down to these clear, precise answers. The use of the square root property is the final, elegant step that unlocks the values of $x$. It’s important to always include the $\pm$ sign because squaring both a positive and a negative number results in a positive number, meaning there are often two roots to a quadratic equation.

Conclusion: Mastering Completing the Square

And there you have it, folks! We've successfully used completing the square to solve for $x$ in the equation $(x+7)(x-9)=25$. We started by expanding and simplifying the equation to get it into standard quadratic form ($x^2 - 2x - 88 = 0$). Then, we skillfully completed the square by adding 1 to both sides, transforming the equation into $(x-1)^2 = 89$. Finally, we applied the square root property to find our two exact solutions: $x = 1 \pm\sqrt{89}$.

This method is incredibly versatile. It's not just for solving equations; it's fundamental to understanding the properties of quadratic functions, like finding the vertex of a parabola and graphing it. The concept of transforming an expression into a squared binomial is a cornerstone of algebra and appears in many other areas of mathematics. Practice is key, guys! The more you work through problems like this, the more intuitive completing the square will become. Remember the steps: expand, simplify, move the constant, add $(b/2)^2$ to both sides, factor, and then use the square root property. You've got this! Keep practicing, and you'll be a completing the square master in no time. Happy solving!