Solving Quadratic Equations by Completing the Square

As noted above, this quadratic does not factor, so I can't solve the equation by factoring. And they haven't given me the equation in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that ready-for-square-rooting form, so I can solve.

First, I put the loose number on the other side of the equation:

Then I look at the coefficient of the x-term, which is −4 in this case. I take half of this number (including the sign), which gives me −2. (I need to keep track of this value. It will simplify my work later on.)

Then I square this value to get +4, and add this squared value to both sides of the equation:

This process creates a quadratic expression that is a perfect square on the left-hand side of the equation. I can factor, or I can simply replace the quadratic with the squared-binomial form, which is the variable, x, together with the one-half number that I got before (and noted that I'd need later), which was −2. Either way, I get the square-rootable equation:

(I know it's a "−2" inside the parentheses because half of −4 was −2. By noting the sign when I'm finding one-half of the coefficient, I help keep myself from messing up the sign later, when I'm converting to squared-binomial form.)

(By the way, this process is called "completing the square" because we add a term to convert the quadratic expression into something that factors as the square of a binomial; that is, we've "completed" the expression to create a perfect-square binomial.)

Now I can square-root both sides of the equation, simplify, and solve:

Using this method, I get the same answer as I had before; namely:

There is one extra step for solving this equation, because the leading coefficient is not 1; I'll first have to divide through to convert the leading coefficient to 1. Here's my process:

Now that I've got all the terms with variables on one side, with the strictly-numerical term on the other side, I'm ready to complete the square on the left-hand side. First, I take the linear term's coefficient (complete with its sign), −(5/2), and multiply by one-half, and square:

Then I add this new value to both sides, convert to squared-binomial form on the left-hand side, and solve:

The two terms on the right-hand side of the last line above can be combined over a common denominator, and this is often ("usually"?) how the answer will be written, especially if the instructions for the exercise included the stipulation to "simplify" the final answer: