1. Jan 5, 2008

### Holocene

Regarding $$\displaystyle{\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$$

How should "$$\displaystyle{\pm}$$" be treated?

I know a square root can be both possitive and negative, but does the quardratic forumla dictate that a possitive root should be added to -b, or does it dictate that a postive root should be subtracted from -b???

2. Jan 5, 2008

### Integral

Staff Emeritus

3. Jan 5, 2008

### Gib Z

Some people get confused because other times they see the plus/minus sign, they have to choose correctly, whilst in this case you do it both. Thats why you may also see the formula written as;

$$x_1= \displaystyle{\frac{-b + \sqrt{b^2 - 4ac}}{2a}}$$

$$x_2 = \displaystyle{\frac{-b - \sqrt{b^2 - 4ac}}{2a}}$$

4. Jan 5, 2008

### Holocene

Sorry, I'm a little confused.

Say a particular solution to a quadratic equation is [tex]\displaystyle{\frac{2 \pm \sqrt{7}}{3}}[\tex]

If 7 was a perfect sqaure, would the root get added to or subtrected from 2?

5. Jan 5, 2008

### Gib Z

It doesn't matter if 7 is a perfect square or not (and it isn't). There are TWO roots to a quadratic equation, hence the quadratic formula has TWO solutions. One of the solutions is when you add the square root part, the other solution is when you subtract it. Look at my previous post.

6. Jan 5, 2008

### Holocene

Okay I've got it. The 2 solutions are adding a possitive root, and subtracting a possitive root, correct? (Subtracting a negative root is = to adding it, adding a negative root is = to subtracting it).

7. Jan 5, 2008

### morphism

$\sqrt{x}$ is used to denote the positive square root of x. Thus $\sqrt{4}=2$ is correct while $\sqrt{4}=\pm2$ is incorrect. Thinking that the latter is correct is too common a mistake - make sure you aren't making it.

8. Jan 5, 2008

### Holocene

Got it, thanks.