Solve the quadratic equation by completing the square.
$x^2-14 x-6=0$
To solve the quadratic equation \(x^2 - 14x - 6 = 0\), we will use the method of completing the square. This technique involves transforming a quadratic expression into a perfect square trinomial plus a constant. Our goal is to isolate \(x\). The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). In this problem, we have \(a = 1\), \(b = -14\), and \(c = -6\).
First, move the constant term to the other side of the equation to isolate the terms involving \(x\):
\[x^2 - 14x = 6\]
To complete the square for the expression \(x^2 - 14x\), we need to add a value that makes the expression a perfect square of the form \((x - d)^2 = x^2 - 2dx + d^2\). Comparing \(x^2 - 14x\) to \(x^2 - 2dx\), we identify \(2d = 14\), so \(d = 7\). The term we need to add is \(d^2 = 7^2 = 49\). We must add this to both sides of the equation to maintain equality:
\[x^2 - 14x + 49 = 6 + 49\]
Now, rewrite the left side of the equation as a perfect square and simplify the right side:
\[(x - 7)^2 = 55\]
To solve for \(x\), take the square root of both sides:
\[x - 7 = \pm \sqrt{55}\]
Finally, isolate \(x\) by adding 7 to both sides:
\[x = 7 \pm \sqrt{55}\]
Thus, the solutions to the equation are \(x = 7 + \sqrt{55}\) and \(x = 7 - \sqrt{55}\).
Completing the square is not just an algebraic exercise; it is a fundamental tool for analyzing physical systems. For example, if you are modeling the trajectory of a ball thrown into the air, its height \(h\) at time \(t\) can often be described by a quadratic equation like \(h(t) = -gt^2 + vt + h_0\). By completing the square, you can rewrite this into vertex form, which immediately reveals the peak height and the time at which that peak occurs. This makes it an essential technique for physics, engineering, and data analysis when you need to find extrema or optimize values in real-world scenarios.