When you're solving quadratics in your homework, you can often get a "hint" as to the "best" method to use, based on the topic and title of the section. For instance, if you're working on the homework in the "Solving by Factoring" section, then you know that you're supposed to solve by factoring. But in the chapter review and on the test, you don't know which section the quadratic came from. Which method should you use?

You could use the Quadratic Formula for everything, but the Formula can be "overkill". For example:

**Solve (***x*+ 1)(*x*– 3) = 0.

This is a quadratic, and I'm supposed to solve it. I could multiply the left-hand side, simplify to find the coefficients, plug them into the Quadratic Formula, and chug away to the answer.

But why would I? I mean, for heaven's sake, this is factorable, and they've already factored it and set it equal to zero for me. While the Quadratic Formula would give me the correct answer, why bother with it? Instead, I'll just solve the factors:

(*x* + 1)(*x* – 3) = 0 *x* + 1 = 0 or *x* – 3 = 0 *x* = –1 or *x* = 3

**The solution is x = –1, 3**

**Solve***x*^{2}– 3*x*– 4 = 0.

This one factors easily:

*x*^{2} – 3*x* – 4 = 0

(*x* + 1)(*x* – 4) = 0 *x* + 1 = 0 or *x* – 4 = 0 *x* = –1 or *x* = 4

**The solution is x = –1, 4**

**Solve***x*^{2}– 4 = 0.

This quadratic has just two terms, and nothing factors out of both, so it's a difference of squares (so I can factor) or it can be reformatted as "(squared part) equals (a number)" so I can square-root both sides. In this case, I can factor:

*x*^{2} – 4 = 0

(*x* + 2)(*x* – 2) = 0 *x* + 2 = 0 or *x* – 2 = 0 *x* = –2 or *x* = 2

**The solution is x = ± 2**

I could have moved the 4 over to the right-hand side of the equation, and then taken the square root of either side of *x*^{2} = 4. This method would have given me the exact same answer as the factoring done above. Use whichever method you prefer.

Find the roots of the quadratic equation: x^{2} + 2x - 15 = 0?

Answer: Option A

Explanation:

x^{2} + 5x - 3x - 15 = 0

x(x + 5) - 3(x + 5) = 0

(x - 3)(x + 5) = 0

=> x = 3 or x = -5.

**Solve***x*^{2}– 7*x*= 0.

This quadratic factors easily:

*x*^{2} – 7*x* = 0 *x*(*x* – 7) = 0 *x* = 0 or *x* – 7 = 0 *x* = 0 or *x* = 7

**The solution is x = 0, 7**