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This module discusses the quadratic formula.

In "Solving Quadratic Equations by Completing the Square" I talked about the common mathematical trick of solving a problem once, using letters instead of numbers , and then solving specific problems by plugging numbers into a general solution.

In the text, you go through this process for quadratic equations in general. The definition of a quadratic equation is any equation that can be written in the form:

${\text{ax}}^{2}+\text{bx}+c=0$

where $a\ne 0$ . By completing the square on this generic equation, you arrive at the quadratic formula:

$x=\frac{-b±\sqrt{{b}^{2}-4\text{ac}}}{2a}$

This formula can then be used to solve any quadratic equation, without having to complete the square each time. To see how this formula works, let us return to the previous problem:

${9x}^{2}-\text{54}x+\text{80}=0$

In this case, $a=9$ , $b=-\text{54}$ , and $c=\text{80}$ . So the quadratic formula tells us that the answers are:

$x=\frac{-\left(-\text{54}\right)±\sqrt{\left(-\text{54}{\right)}^{2}-4\left(9\right)\left(\text{80}\right)}}{2\left(9\right)}$

We’ll use a calculator here rather than squaring 54 by hand....

$x=\frac{\text{54}±\sqrt{\text{2916}-\text{2880}}}{\text{18}}=\frac{\text{54}±\sqrt{\text{36}}}{\text{18}}=\frac{\text{54}±6}{\text{18}}=\frac{9±1}{3}$

So we find that the two answers are $\frac{\text{10}}{3}$ and $\frac{8}{3}$ , which are the same answers we got by completing the square.

Using the quadratic formula is usually faster than completing the square, though still slower than factoring. So, in general, try to factor first: if you cannot factor, use the quadratic formula.

So why do we learn completing the square? Two reasons. First, completing the square is how you derive the quadratic formula. Second, completing the square is vital to graphing quadratic functions, as you will see a little further on in the chapter.

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