When you set a quadratic polynomial equal to 0, you get a quadratic equation. Quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where a, b, and c are constants (real numbers) and a ≠ 0, and x is a variable that you have to solve for.
For example, 2x2 + 3x + 2 = 0 is a quadratic equation while 3x + 2 is not a quadratic equation.
There are two ways of solving a quadratic equation:
The simplest way to solve a quadratic equation is to try to factor the equation into two binomials. It is the process of breaking apart of an equation into factors (or separate terms) such that when the separate terms are multiplied together, they produce the original equation.
For example, x2 - x - 2 = (x + 1)(x - 2)
In this case, the equation x2 - x - 2 = 0 can be broken apart into two factors: (x + 1)(x - 2) = 0; such that when these two separate terms are multiplied together, the result is the original equation.
For example, x2 - 6x + 5 = 0
To factor this trinomial, consider what numbers multiply together to become 5 that also have a sum of –6. The two factors of 5 are 5 and 1 or –5 and –1. To get a sum of –6, you need to go with the negative values. Doing so gives these two binomial factors: (x – 5) and (x – 1). So the resulting equation is (x – 5)(x – 1) = 0.
To solve for x, you set each of the binomial factors equal to 0. You can do so because you know that one of the factors must equal 0 if their product is 0.
x – 5 = 0
x = 5
x – 1 = 0
x = 1
Now, the solutions (or roots) to the equation are: x = 1 and x = 5. Both 1 and 5 are possible solutions for x in this quadratic equation.
Finding the solution set for a quadratic equation made up of the difference of perfect squares (like x2 - y2) is simple if you remember that:
x2 - y2 = (x - y)(x + y)
If the GMAT presents you with the task of solving for x in an equation where the difference of perfect squares is equal to 0, you know that x equals the positive and negative values of the square root of y2 (which is the second term).
So, if you were to find the solution set for x2 - 49 = 0, you determine the square root of the second term (49), which is 7. The factors, then, are (x + 7) and (x – 7). Therefore, the solution set for this problem is x = –7 and x = 7, which are the positive and negative values of the second term’s square root.
Root is a solution to an equation such that f(x) = 0. For example, for the quadratic equation x2 + x - 12 = 0, x = 3 and x = -4 are both roots. Quadratic equations usually have two possible solutions.
While the quickest means to solve a quadratic equation is often through factoring, a quadratic equation can be too complex to easily factor or it simply does not factor. In these cases, the quadratic formula can be used as it will always find the correct answer, regardless of the properties of the equation.
For a quadratic equation, which has the form ax2 + bx + c = 0, the roots are given by the formula: