## Solving Quadratic Equations – Choosing an Efficient Technique

While learning to solve quadratic equations, some students decide to always use the quadratic formula because they can use it for any equation. But, depending on the equation, there are often more efficient techniques. In this blog I’ll go over some of the other techniques and help you to determine when to use them.

Solving by Factoring

This should be the first tool to try when solving a quadratic equations that is in standard form: $ax^2+bx+c$. If you can factor the quadratic expression, determining the solutions is quite easy.

$x^2+5x=2x+28$
$x^2+3x-28=0$  Rewrite in standard form.
$(x+7)(x-4)=0$  Factor the quadratic expression.
$\begin{array}{rcl} x+7=0 & or & x-4=0 \end{array}$ Set each factor equal to 0.
$\begin{array}{rcl} x=-7 & or & x=4 \end{array}$ Solve.

In most cases, if you can factor the expression you can determine the solutions by inspection, which speeds up the process.

I tell my students to apply the “10 second rule” when trying to factor the expression – if you cannot factor the expression quickly (10 seconds or less) then you should move on to another technique. This is especially true when the leading coefficient a is not equal to 1.

Solving by Extracting Square Roots

This technique, which involves taking the square root of both sides of an equation, should be used when an equation contains one squared term and constants. Here are some examples:
$x^2=40 \\ x^2+17=-19 \\ (x+7)^2=-45 \\ (x-2)^2-15=9$

The steps for this technique are:

1. Isolate the squared term.
2. Take the square root of each side of the equation. Use the $\pm$ sign when taking the square root of the constant term.
3. Solve the resulting equation.

Here’s an example worked out for the equation $(x-4)^2+11=-17$.

$(x-4)^2+11=-17$
$(x-4)^2=-28$  Isolate the squared term by subtracting 11.
$\sqrt{(x-4)^2}=\pm \sqrt{-28}$  Take the square root of both sides.
$x-4=\pm 2i\sqrt7$  Simplify each square root.
$x=4\pm 2i\sqrt7$  Solve for x by adding 4 to both sides.

Solving by Completing the Square

Solving by completing the square works well whenever the leading coefficient is 1 and the coefficient of the “x-term” is even. If the leading coefficient is not 1, you must divide through by the leading coefficient first and this often introduces fractions to the process. The same thing occurs when the coefficient of the x-term is odd.

Here is the procedure.

1. Isolate the variable terms.
2. Add $(\frac b2)^2$ to both sides of the equation.
3. Factor the resulting trinomial.
4. Solve by extracting square roots

Here’s an example.

$x^2+8x+11=0$
$x^2+8x=-11$  Isolate the variable terms.
$x^2+8x+16=-11+16$ Add $(\frac82)^2=4^2=16$ to both sides.
$(x+4)^2=5$  Factor the trinomial.
$\sqrt{(x+4)^2}=\pm \sqrt5$  Take the square root of both sides.
$x+4=\pm \sqrt5$  Simplify the square root.
$x=-4\pm \sqrt5$  Solve for x by subtracting 4 from both sides.

Solving by Using the Quadratic Formula

Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ only when none of the previous techniques can be applied efficiently. The quadratic formula can be used to solve any equation, but is often not the best choice. If you have a dirty window, a sledgehammer will clear the view of dirt but it is not the best tool. The same is true for the quadratic formula.

Summary

I hope that you find this helpful. If you have any questions about solving quadratic equations, or if you have any other topics you are interested in, you can reach me through the contact page at my web site – georgewoodbury.com.

-George

I am a math instructor at College of the Sequoias in Visalia, CA. Each Wednesday I post an article related to General Teaching on my blog. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website – http://georgewoodbury.com.

• 1. Diane Bauman  |  October 27, 2010 at 12:11 pm

I love your 10-second rule. Good one to keep in mind!

I always go through the pro’s and con’s of each method and let them choose their method (but they have to show that they CAN do each one first). Its amazing to me how many will choose “completing the square’ even with an odd middle-coefficient. There’s no accounting for taste! ha.

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