A higher order equation, which is also called a quadratic equation, is where at least one of the variables is raised to something other than the power of 1. The following are all quadratic, or higher order, equations:
-3 + x2 = 22
4y3 + 3z = 36 - z
x1.5 = 3
n2.3 / n0.3 = 16
Solving higher order equations is just a little bit more involved than solving linear equations, and so expect to see such topics appear on the more difficult GRE questions. You still use the same concepts of adding, subtracting, multiplying, or dividing both sides of the equation by the same number or variable, but you may also be required to do additional manipulations. One such manipulation requires you to rearrange the equation so that one side is equal to 0, and then allows you to perform factoring and solve for the needed variable. It's best to see how this works by example, and here we solve the first equation, and on the next page we solve the fourth equation. Notice how we need to use the rules of exponentiation in some cases. For the first problem, you might be tempted to do the following (which is partially true):
| Step 0 | Original Equation, solve for x | -3 + x2 = 22 |
| Step 1 | Add 3 to each side | x2= 25 |
| Step 2 | Take the square root of both sides | x = 5 |
It seems that x is equal to 5, right? Wrong! Yes, when x is equal to 5 then the equation is true, but notice that when x is equal to -5, then the equation is also true, because (-5)2 = 25, so x can be one of two values and the equation is still valid. So how do we determine this? Simple. We rearrange the equation so that one side is equal to 0, and then we factor. So here is the same problem done the more complete way:
| Step 0 | Original Equation, solve for x | -3 + x2 = 22 |
| Step 1 | Subtract 22 from each side | -25 + x2 = 0 |
| Step 2 | Rewrite the left side to make the problem easier to manipulate | x2 - 25 = 0 |
| Step 3 | Using the difference of squares rule, rewrite the equation | (x - 5)(x + 5) = 0 |
| Step 4 | In order to have the left hand side equal zero and the equation true, we use the fact that any number multiplied by 0 is 0, and we know that one of the parts of the left hand side must be zero, so when x = 5. Thus, the first bracket of the left hand side is 0, and when x = -5, then the second bracket of the left hand side is 0, and that is how we get our two possible solutions. | x = 5 x = -5 |