Completing The Square: Solving $x^2 - 14x - 6 = 0$

Quadratic equations are a fundamental part of algebra, and one of the most elegant methods for solving them is by completing the square. This technique allows us to transform a standard quadratic equation into a form where we can easily isolate the variable. Today, we're going to tackle the specific equation $x^2 - 14x - 6 = 0$ using this powerful method.

Understanding the Goal of Completing the Square

Before we dive into the steps, let's get a clear picture of what completing the square actually means. The core idea is to manipulate a quadratic expression of the form $ax^2 + bx + c$ into a perfect square trinomial, which looks like $(x+h)^2$ or $(x-h)^2$. A perfect square trinomial is a trinomial that can be factored into the square of a binomial. For example, $x^2 + 6x + 9$ is a perfect square trinomial because it factors into $(x+3)^2$. Similarly, $x^2 - 10x + 25$ factors into $(x-5)^2$. The beauty of this is that if we can get our quadratic equation into the form $(x+h)^2 = k$, we can then take the square root of both sides and easily solve for $x$. This method is particularly useful when the quadratic expression doesn't factor easily using traditional methods.

Our target equation is $x^2 - 14x - 6 = 0$. Notice that the terms involving $x$ are $x^2 - 14x$. We want to add a constant term to this expression to make it a perfect square trinomial. Let's think about the general form of a squared binomial: $(x+h)^2 = x^2 + 2hx + h^2$. If we compare this to our expression $x^2 - 14x$, we can see that the coefficient of the $x$ term, $2h$, must be equal to $-14$. This gives us a way to find the value of $h$. By setting $2h = -14$, we can solve for $h$ by dividing both sides by 2, which gives us $h = -7$.

Now that we know $h = -7$, we can determine the constant term needed to complete the square. According to the formula $(x+h)^2 = x^2 + 2hx + h^2$, the constant term is $h^2$. So, for our case, the constant term we need is $(-7)^2$. Calculating this, we get $(-7)^2 = 49$. Therefore, if we add 49 to $x^2 - 14x$, we will get a perfect square trinomial: $x^2 - 14x + 49$. This trinomial can be factored as $(x-7)^2$.

Step-by-Step Solution for $x^2 - 14x - 6 = 0$

Let's now apply this understanding to solve our specific equation, $x^2 - 14x - 6 = 0$. The first step in solving by completing the square is to isolate the terms containing $x$ on one side of the equation. To do this, we need to move the constant term, $-6$, to the right side of the equation. We achieve this by adding 6 to both sides:

$x^2 - 14x - 6 + 6 = 0 + 6$

This simplifies to:

$x^2 - 14x = 6$

Now, we need to complete the square on the left side of the equation. As we determined earlier, the coefficient of our $x$ term is $-14$. To find the number we need to add to complete the square, we take half of this coefficient and square it. Half of $-14$ is $-7$, and squaring $-7$ gives us $(-7)^2 = 49$.

Crucially, whatever we add to one side of the equation, we must also add to the other side to maintain equality. So, we add 49 to both sides of our equation:

$x^2 - 14x + 49 = 6 + 49$

The left side, $x^2 - 14x + 49$, is now a perfect square trinomial. We can factor it as $(x - 7)^2$. The right side simply adds up to $6 + 49 = 55$. So, our equation now looks like this:

$(x - 7)^2 = 55$

This is the form we were aiming for! The variable $x$ is now contained within a squared term, and we can proceed to solve for $x$ by taking the square root of both sides. Remember that when taking the square root of a number, there are two possible results: a positive root and a negative root.

$\sqrt{(x - 7)^2} = \pm\sqrt{55}$

This simplifies to:

$x - 7 = \pm\sqrt{55}$

Our final step is to isolate $x$ by adding 7 to both sides of the equation:

$x = 7 \pm\sqrt{55}$

This gives us two distinct solutions for the quadratic equation $x^2 - 14x - 6 = 0$:

$x_1 = 7 + \sqrt{55}$

$x_2 = 7 - \sqrt{55}$

These are the exact solutions. If an approximate decimal answer is needed, you would calculate the approximate value of $\sqrt{55}$ (which is about 7.416) and substitute it.

Why is Completing the Square Important?

The method of completing the square is not just a way to solve quadratic equations; it's a foundational technique that underpins other important mathematical concepts. For instance, it's used to derive the quadratic formula itself. The quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which provides a direct solution for any quadratic equation in the form $ax^2 + bx + c = 0$, is derived by applying the completing the square method to the general form of the quadratic equation. Understanding completing the square provides deeper insight into why the quadratic formula works and its structure.

Furthermore, completing the square is essential when working with conic sections, such as circles, ellipses, and hyperbolas. The standard equations for these curves often involve squared terms of $x$ and $y$, and rewriting them into a standard form to identify their properties (like the center and radius of a circle) frequently requires the completing the square technique. For example, to find the center $(h, k)$ and radius $r$ of a circle given by the equation $x^2 + y^2 + Dx + Ey + F = 0$, you would rearrange the terms and complete the square for both the $x$ and $y$ variables.

This method also has applications in calculus, particularly when dealing with integration. Certain integrals can be simplified or made solvable by transforming the integrand using completing the square. It's a versatile algebraic tool that demonstrates how different parts of mathematics are interconnected. By mastering completing the square, you're building a stronger foundation for tackling more complex mathematical problems across various fields.

In summary, solving $x^2 - 14x - 6 = 0$ by completing the square involved isolating the $x$ terms, adding the square of half the $x$-coefficient to both sides to create a perfect square trinomial, factoring that trinomial, and then solving for $x$ by taking the square root. This process yielded the solutions $x = 7 \pm\sqrt{55}$.

For more on quadratic equations and solving techniques, you can explore resources from ** Khan Academy** or ** Math is Fun**.