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The Math Thread (4 N3rdS!!!) (pg. 6)
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| ierxium |
(3x-1)(2x+1)
Someone want to double check that? :conf: |
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| stevieboy32808 |
| quote: | Originally posted by ierxium
(3x-1)(2x+1)
Someone want to double check that? :conf: |
Sorry bro, that's not it. The answer is a couple posts back... |
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| ierxium |
| quote: | Originally posted by stevieboy32808
Sorry bro, that's not it. The answer is a couple posts back... |
I know. I was talking about the new problem. Post above mine. |
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| ::TranceVanDyk:: |
| quote: | Originally posted by ierxium
(3x-1)(2x+1)
Someone want to double check that? :conf: |
THANK YOU!!
i was having a bit of brain lapse. |
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| ierxium |
| quote: | Originally posted by ::TranceVanDyk::
THANK YOU!!
i was having a bit of brain lapse. |
What's my prize? |
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| stevieboy32808 |
| quote: | Originally posted by ierxium
(3x-1)(2x+1)
Someone want to double check that? :conf: |
What a nooben I am. yes you're right.:D |
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| Floorfiller |
since this is the math nerd thread...
does anyone have any good suggestions for books on set theory? or related math topics... |
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| ::TranceVanDyk:: |
dealin with imaginaries now, i know the answers, but im having a little fun.
find the zero's of x^3 + x^2 - 4x - 24
HINT: its long, so i'll give u this. (x-3) is a factor. |
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| stevieboy32808 |
x^2(x+1)-4(x+6)
So the zeros are 0, -1, and -6. please verify this TranceVanDyk.... |
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| ::TranceVanDyk:: |
| quote: | Originally posted by stevieboy32808
x^2(x+1)-4(x+6)
So the zeros are 0, -1, and -6. please verify this TranceVanDyk.... |
nope, im headin out of work, ill give u the answer when i get home. til then;) try try again. |
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| ::TranceVanDyk:: |
| quote: | Originally posted by ierxium
What's my prize? |
my respect;)
| quote: | x^2(x+1)-4(x+6)
So the zeros are 0, -1, and -6. please verify this TranceVanDyk.... |
the answers are x = -2(+/-)2i, 3
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find the zero's of x^3 + x^2 - 4x - 24
three roots. u find out through descartes rule that there is 1 positive zero, and 2 imaginary ones. so u know there are going to be imaginaries.
the hint was, (x-3) was a factor of the equation above. knowing that, u divide the equation by (x-3). the zero which is 3 b/c (3-3=0) is divided by the equation to get...
x^2 + 4x + 8
u cant factor it. so you use the quadratic equation to find the zeros. your answer should then be -2(+/-)2i.
answers are then x = 3, -2(+/-)2i
if u checked your answers 0, -1, and -6, they would not have equaled 0 in the original equation. |
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