2.1 Solving quadratic equations using the method of completing the

Overview

• The Logic Behind The Method
• The Method Of Completing The Square

The logic behind the method

Suppose we wish to solve the quadratic equation $x^2-3x-1=0.$ Since the equation is not of the form $x^2=K,$ we cannot use extraction of roots. Next, we try factoring, but after a few trials we see that $x^2-3x-1$ is not factorable. We need another method for solving quadratic equations.
The method we shall study is based on perfect square trinomials and extraction of roots. The method is called solving quadratic equations by completing the square . Consider the equation $x^2+6x+5=0.$
This quadratic equation could be solved by factoring, but we’ll use the method of completing the square. We will explain the method in detail after we look at this example. First we’ll rewrite the equation as

$$x^2+6x=-5$$

Then, we’ll add 9 to each side. We get

$$x^2+6x+9=-5+9$$

The left side factors as a perfect square trinomial.

$${\left(x+3\right)}^2=4$$

We can solve this by extraction of roots.
$$\begin{array}{lllllll}x+3\hfill & =\hfill & +\sqrt{4}\hfill & \hfill & \hfill & \hfill & \hfill \\ x+3\hfill & =\hfill & \pm 2\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & \pm 2-3\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & +2-3\hfill & \hfill & \text{and}\hfill & \hfill & x=-2-3\hfill \\ \hfill x& =\hfill & -1\hfill & \hfill & \text{and}\hfill & \hfill & -5\hfill \end{array}$$

Notice that when the roots are rational numbers, the equation is factorable.
The big question is, “How did we know to add 9 to each side of the equation?” We can convert any quadratic trinomial appearing in an equation into a perfect square trinomial if we know what number to add to both sides. We can determine that particular number by observing the following situation:

Consider the square of the binomial and the resulting perfect square trinomial

$${\left(x+p\right)}^2=x^2+2px+p^2$$

Notice that the constant term (the number we are looking for) can be obtained from the linear term $2px.$ If we take one half the coefficient of $x,\frac{2p}{2}=p,$ and square it, we get the constant term $p^2.$ This is true for every perfect square trinomial with leading coefficient 1.

In a perfect square trinomial with leading coefficient 1, the constant term is the square of one half the coefficient of the linear term.

Study these examples to see what constant term will make the given binomial into a perfect square trinomial.