set to zero |
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 Sep 30 2008, 08:55 PM
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#1
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 Death is a promise given to us at birth  Group: Official Designer Posts: 4,757 Joined: Mar 2004 Member No: 7,459  |
i need help with this problem. I can't seem to do it. I need to solve for "x".
Help immediately. 
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Sep 30 2008, 09:43 PM
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#2
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 /人◕‿‿◕人\ Group: Official Member Posts: 8,283 Joined: Dec 2007 Member No: 602,927 |
4x^3-6x^2+1=0
4x^3-6x^2=-1 10x^5=-1 50x=-1 x=-50 I could very easily be wrong. Exponents screw me up quite often. Edit: Found my calculator, I was way off. Edit: Go with no solution, above poster is smart. |
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Sep 30 2008, 09:46 PM
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#3
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 in a matter of time Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 |
^ That is completely, utterly, wrong. On so many freakin' levels.
Factored: (2x-1)(2x^2-2x-1) = 0 One solution is x = 0.5. Use the quadratic formula to solve for (2x^2-2x-1) = 0 to find the other two values of x that make the whole thing = 0. |
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Sep 30 2008, 09:48 PM
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#4
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 Group: Staff Alumni Posts: 7,155 Joined: Feb 2005 Member No: 95,404 |
^x2 Yeah, that's not right. You can't just add x^3 and x^2 together.
Yayuh Gigi, unleash the math geek in you! |
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Sep 30 2008, 09:50 PM
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#5
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in a matter of time Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 |
Actually if you have a graphing calculator (or if you're allowed to use one), I'd probably just graph it and then find the zeros. Factoring cubics is NOT fun.
XTC, you could say that you committed every single algebra crime you could think of. The mathematical portion of my brain died a little reading that. |
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Sep 30 2008, 09:59 PM
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#6
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 and so it is Group: Human Posts: 1,304 Joined: Feb 2004 Member No: 3,085 |
1) type it into y=
2) graph it 3) 2nd -> trace -> zero 4) find left bound and press enter 5) find right bound and press enter 6) guess where it is approx (doesnt matter) 7) x = .5 ; y = 0 and if you plug .5 as "x" into your equation, it should equal 0. |
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Sep 30 2008, 11:00 PM
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#7
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Death is a promise given to us at birth Group: Official Designer Posts: 4,757 Joined: Mar 2004 Member No: 7,459 |
everyone could use a calculator, but that's not how you learn.
how do you factor cubic? |
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Sep 30 2008, 11:06 PM
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#8
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in a matter of time Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 |
That's true and all, but factoring cubics is honestly the dumbest thing ever. You wouldn't be sacrificing much if you did use a calculator.
Factoring cubics involves finding a common factor within all the trinomial. It's a little complicated and not worth learning, unless you're being tested on it, in which case you should have learned how. I honestly learned this in 10th grade for about 5 minutes and have no idea how to do it anymore. However, there are plenty of tutorials online, just search "factoring cubics" on Google. |
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Sep 30 2008, 11:07 PM
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#9
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Group: Staff Alumni Posts: 7,155 Joined: Feb 2005 Member No: 95,404 |
Yeah, I remember learning this a couple years ago, but right now I'm just kind of going
 at factoring cubic functions.
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Sep 30 2008, 11:34 PM
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#10
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and so it is Group: Human Posts: 1,304 Joined: Feb 2004 Member No: 3,085 |
so i dug up old precalc/trig notes cause i remembered i did something like this with a quadnomial... but THEN i realized that we were also given an irrational solution, which is not related to your question at all which also makes this post completely pointless and void.
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