2.5 Quadratic equations  (Page 3/14)

Using the square root property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property    , in which we isolate the $x^2$ term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the $x^2$ term so that the square root property can be used.

Completing the square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation    known as completing the square    . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a , must equal 1. If it does not, then divide the entire equation by a . Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example $x^2+4x+1=0$ to illustrate each step.

1. Given a quadratic equation that cannot be factored, and with $a=1,$ first add or subtract the constant term to the right sign of the equal sign.

   $$x^2+4x=\mathrm{-1}$$

2. Multiply the b term by $\frac{1}{2}$ and square it.

   $$\begin{array}{ccc}\hfill \frac{1}{2}(4)& =& 2\hfill \\ \hfill 2^2& =& 4\hfill \end{array}$$

3. Add ${\left(\frac{1}{2}b\right)}^2$ to both sides of the equal sign and simplify the right side. We have

   $$\begin{array}{ccc}\hfill x^2+4x+4& =& -1+4\hfill \\ \hfill x^2+4x+4& =& 3\hfill \end{array}$$

4. The left side of the equation can now be factored as a perfect square.

   $$\begin{array}{ccc}\hfill x^2+4x+4& =& 3\hfill \\ \hfill {(x+2)}^2& =& 3\hfill \end{array}$$

5. Use the square root property and solve.

   $$\begin{array}{ccc}\hfill \sqrt{{(x+2)}^2}& =& \pm \sqrt{3}\hfill \\ \hfill x+2& =& \pm \sqrt{3}\hfill \\ \hfill x& =& \mathrm{-2}\pm \sqrt{3}\hfill \end{array}$$

6. The solutions are $x=\mathrm{-2}+\sqrt{3},$ $x=\mathrm{-2}-\sqrt{3}.$