• Aug 11th 2009, 09:34 AM
using $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

With:

a = 1, b = (-4) & c = (-3)

What do we do when we need to square root the -4 ?

using $\displaystyle x=\frac{-(-4)\pm\sqrt{-4}}{2}$
• Aug 11th 2009, 09:45 AM
skeeter
Quote:

using $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

With:

a = 1, b = (-4) & c = (-3)

What do we do when we need to square root the -4 ?

using $\displaystyle x=\frac{-(-4)\pm\sqrt{-4}}{2}$

$\displaystyle b^2 - 4ac$

$\displaystyle (-4)^2 - 4(1)(-3) = 16 + 12 = 28$ , not $\displaystyle -4$
• Aug 11th 2009, 09:53 AM
$\displaystyle x=\frac{-(-4)\pm\{{5.921}}{2}$

?
• Aug 11th 2009, 09:59 AM
skeeter
Quote:

$\displaystyle x=\frac{-(-4)\pm\{{5.921}}{2}$

?

$\displaystyle \frac{4 \pm \sqrt{4 \cdot 7}}{2}$

$\displaystyle \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$
• Aug 11th 2009, 10:11 AM
Leslon
First of all, if you evaluate $\displaystyle x=b^2-4ac$, with the numbers you provided, you end up with $\displaystyle \sqrt{16-(-12)} = \sqrt{28}$ . So no problem here, i guess.

In the other case of a square root of a negative number, you will have to work with Complex numbers. Let us suppose you have to solve $\displaystyle x=\frac{-(-4)\pm\sqrt{-4}}{2}$. The next step is to write $\displaystyle \sqrt{-4}$ as $\displaystyle \sqrt{4i^2}$. If solve this root, you get $\displaystyle 2i$. Now we have $\displaystyle x=\frac{-(-4)\pm2i}{2}$
. Let's simplify all that and we get $\displaystyle x=2\pm i$

Greets,
Leslon