## Quadratic Formula on the New SAT

Based on the official tests we have at our disposal, it seems that there aren’t too many questions that require the use of the quadratic formula. But…and this is an important ‘but’, if test day you face one or two questions that can only be solved using the quadratic formula, not even having a chance at these questions can hurt you psychologically. So though two points might seem miniscule in the overall scheme that psychological effect can reverberate throughout the rest of the math section.

Okay, my intention here is not to scare you. It is to give you the necessary tools. In this case, simply memorize the quadratic formula, do a few practiced questions using this concept, and test day you won’t be bullied by this concept.

## Why is the quadratic formula important?

On the old SAT and in many introductory algebra classes, you are taught the FOIL method. The second-degree polynomials you end up seeing can, unsurprisingly, be solved using FOIL.

x^{2}-4x-5 = 0

This equation can be solved as follows:

(x-5)(x+1) = 0

In solving this, I wanted to make sure that the product of the last numbers in parenthesis is -5 and that the sum of these numbers is -4. In each of these FOIL cases, the product and the sum, and indeed the solution or solutions to x, will be integers.

In other words, if the equation is in the form **ax +bx+c**, a, b, c and will all equal integers and the solutions to x will both be integers.

But when do you know to use the quadratic formula? One give away is when the answer choices have a radical sign and a plus or minus sign. Something like this:

4±√2

Another indication is when the product of two integers can not result in b, as in the equation ax^2+bx+c.For instance,take a look at the following equation:

x^{2}-5x-5 = 0

In this case, there are no integers whose product is -5 and that sum to -5. For instance, the only factors of -5 are 1/ -1 and 5/-5. None of those will sum to -5. So in order to find out which two numbers, when multiplied equal -5, but when added equal -5, we have to use the quadratic formula.

**Now we MUST use the quadratic formula.** First, though, I’ve provide the typical polynomial expression so you can easily refer to what a, b, and c correspond to.

ax^{2}+bx+c

Quadratic formula:

Now let’s solve x^{2}-5x-5=0

Here a corresponds to 1, (when you have x^{2} by itself, a equals 1), b corresponds to -5, and c corresponds to -5, too.

Plugging those values into the quadratic formula we get:

[-5±√(25+20)]/2

(-5±3√5)/2