G
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Help me please, I need to know the work for it and an explanation on how I got it, and I'm really confused. :?
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G'day, Ambeezy.
If we want to solve \(\displaystyle x^2 \, + \, 2ix \, - \, 4 \, = \, 0\) fox x, we can apply the quadratic formula as we would usually.
\(\displaystyle \L x = \frac{-b \, \pm \, \sqrt{b^2 \, - \, 4ac}}{2a}\)
Here: a = 1, b = 2i, c = -4.
Plug them in and remember that \(\displaystyle i^2 = -1\).
See how you go.
If we want to solve \(\displaystyle x^2 \, + \, 2ix \, - \, 4 \, = \, 0\) fox x, we can apply the quadratic formula as we would usually.
\(\displaystyle \L x = \frac{-b \, \pm \, \sqrt{b^2 \, - \, 4ac}}{2a}\)
Here: a = 1, b = 2i, c = -4.
Plug them in and remember that \(\displaystyle i^2 = -1\).
See how you go.
Hi,
I am trying to practice for a math placement test I have coming up and I came across this problem. I found a post with this problem already but it could only help me so much so far I can get up to:
If we want to solve \(\displaystyle x^2 \, + \, 2ix \, - \, 4 \, = \, 0\) fox x, we can apply the quadratic formula as we would usually.
\(\displaystyle \L x = \frac{-b \, \pm \, \sqrt{b^2 \, - \, 4ac}}{2a}\)
Here: a = 1, b = 2i, c = -4.
Plug them in and remember that \(\displaystyle i^2 = -1\).
Now I understand all of this, but I can't seem to get past:
\(\displaystyle \L x = \frac{-2i \, \pm \, \sqrt{12}}{2}\)
If someone could walk me through the rest of this process, it would be much appreciated.
Thanks,
dkarp520
I am trying to practice for a math placement test I have coming up and I came across this problem. I found a post with this problem already but it could only help me so much so far I can get up to:
If we want to solve \(\displaystyle x^2 \, + \, 2ix \, - \, 4 \, = \, 0\) fox x, we can apply the quadratic formula as we would usually.
\(\displaystyle \L x = \frac{-b \, \pm \, \sqrt{b^2 \, - \, 4ac}}{2a}\)
Here: a = 1, b = 2i, c = -4.
Plug them in and remember that \(\displaystyle i^2 = -1\).
Now I understand all of this, but I can't seem to get past:
\(\displaystyle \L x = \frac{-2i \, \pm \, \sqrt{12}}{2}\)
If someone could walk me through the rest of this process, it would be much appreciated.
Thanks,
dkarp520
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle \frac{-2i\pm\sqrt{12}}{2} \ = \ \frac{-2i\pm2\sqrt3}{2} \ = \ -i\pm\sqrt3\)