Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I factored -2x2 - 8x + 16Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
Last edited by Roc18 (2008-10-16 17:41:17)
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
Last edited by VicktorVauhn (2008-10-16 17:42:04)
I think he's right.VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
what happened to the other X? it magically turned into a -1?VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
He factored it out with the sqrt 3 I believeHurricane2k9 wrote:
what happened to the other X? it magically turned into a -1?VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:
Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
You guys make my head hurt x(sqrt(3)-1) is x*(sqrt(3)-x, x times -1 = -xHurricane2k9 wrote:
what happened to the other X? it magically turned into a -1?VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:
Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
thisVicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
He factored out the x from the sqrt(3) and the x, which turned into x(sqrt(3) -1)Hurricane2k9 wrote:
what happened to the other X? it magically turned into a -1?VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:
Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
solving the equation doesn't have a damn thing to do with a square root, its exactly the same if 5, 10, C, A, W, or a smiley face was in its place.Hurricane2k9 wrote:
Fuck square roots
except square roots can't be in the denominator dumbass.VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
Don't do that, you may introduce new solutions. If you do do it this way, then you have to test your results in the original equation to make sure it holds.Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
lol +1aj0404 wrote:
except square roots can't be in the denominator dumbass.VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:
Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
This is the correct way of doing it.VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
Yeh, I was completely wrong when I did that, checked my answers. I guess this is what happens when you just came back from 2 lectures and have only had a few hours of sleep. Eh.Vub wrote:
Don't do that, you may introduce new solutions. If you do do it this way, then you have to test your results in the original equation to make sure it holds.Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.3X2 = X2 - 8X + 16
I think.
Why not? It will work plenty fine I promise. That's just a stupid rule to annoy students in basic math classes. Once your past Algebra class no one gives a damn.aj0404 wrote:
except square roots can't be in the denominator dumbass.VicktorVauhn wrote:
why the fuck would you square both sides when x isn't even under a radical? square root of 3 is just a number like anything else, if this was 5x = x-4 would you square it?Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:
Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.
I think.
x*sqrt(3)=x-4
x*sqrt(3) -x=-4
x*(sqrt(3)-1)=-4
x=-4/(sqrt(3)-1)
done. Don't need quadratic formula and squaring stuff...Its really basic algebra...
Haha yeah I reckon, I learnt that the hard way...I'll never forget that maths test *cries*Smithereener wrote:
Yeh, I was completely wrong when I did that, checked my answers. I guess this is what happens when you just came back from 2 lectures and have only had a few hours of sleep. Eh.Vub wrote:
Don't do that, you may introduce new solutions. If you do do it this way, then you have to test your results in the original equation to make sure it holds.Smithereener wrote:
If I'm not mistaken, you do square both sides. Thus you end up with:
Then solve for X via factoring or quadratic formula. You'll probably have to use the quadratic.
I think.
Oh well, I guess at least we can all agree to laugh at Hurricane.Vub wrote:
Haha yeah I reckon, I learnt that the hard way...I'll never forget that maths test *cries*Smithereener wrote:
Yeh, I was completely wrong when I did that, checked my answers. I guess this is what happens when you just came back from 2 lectures and have only had a few hours of sleep. Eh.Vub wrote:
Don't do that, you may introduce new solutions. If you do do it this way, then you have to test your results in the original equation to make sure it holds.
yes, but it sounds like he's in algebra class to me. and i just felt like being a dick.VicktorVauhn wrote:
Why not? It will work plenty fine I promise. That's just a stupid rule to annoy students in basic math classes. Once your past Algebra class no one gives a damn.
I'm not in a 'basic' math class. I'm in grade 11, so I'm in Math 20 Pure Advanced. I'm taking 30 level math next semester.VicktorVauhn wrote:
Why not? It will work plenty fine I promise. That's just a stupid rule to annoy students in basic math classes. Once your past Algebra class no one gives a damn.
