lol?SandWraith wrote:Complete the square.
For real solutions, a= -6+sqrt(12), -6-sqrt(12)
b=+sqrt(-2+sqrt(10)), -sqrt(-2+sqrt(10))
If you wanted help, just ask. You always try to sound so elevated on here, and I don't know why.
[a^2 + 12a + 24 = 0] =>Switch Kicker wrote:Factor the fallowing equation.
a^2 + 12a + 24 = 0
Solve for a.
Try this one too.
2b^4 + 8b^2 - 12 = 0
Solve for b.
This is just cuz I'm bored so I thought I'd grab a few basic algebra questions, lol. If I don't have a response in a few hours, I'll post a hint.
I don't expect this to be hard, lol. This is by no means college level math, lol.
Peace,
Fred.
(a + 3)] =>(a + 3) = 0] Glad you like my avatar.Switch Kicker wrote: [a^2 + 12a + 24 = 0] =>
[a^2 + 12a + 24] =>
[(a +
[(a +
a = -8, -3
Easy.
Try plugging those back into the equation.
For a=-8, a^2 + 12a +24 = 64 - 96 + 24 = -8 which does not equal 0
For a=-3, a^2 + 12a + 24 = 9 - 36 + 24 = -3 which does not equal 0
[2b^4 + 8b^2 - 12 = 0] =>
[2b^4 + 8b^2 - 12] =>
[(2b^2 + 6)(2b^2 - 2)] => that gives a coefficient of +4 for the b^2 term, not +8 as required
[(2b^2 + 6)(2b^2 - 2) = 0]
(2b^2 -2) = 0
b = 1
Again, plug it in. 2(1^4) + 8(1^2) - 12 = 2 + 8 - 12 = -2 which does not equal 0. Same thing for the next one
(2b^2 + 6) = 0
b = -1.73 (Rounded off, lol. It's the -square root of 3.)
If not, I'm gonig to feel really awkward, edit all my posts to be empty, and then go cry in a corner in shame of my rediculous lack of attention to detail, lol. True, that would give +8b^2 as you stated, I didn't notice that you had a 2 in the parentheses for each b^2. The problem with that, though, is that 2b^2 * 2b^2 = 4b^4 and you want 2b^4. The fact that plugging your solution back into the equation doesn't yield 0 is a giveaway that there was an error somewhere. It's a useful technique when the equations get really hairy.Switch Kicker wrote: (2b^2 + 6)(2b^2 - 2)
6 multiplied by -2 = -12
(2b^2 + 6)(2b^2 - 2)
6 multiplied by 2b^2 = 12b^2
(2b^2 + 6)(2b^2 - 2)
-2b^2 multiplied by -2 = -4b^2.
12b^2 + -4b^2 = 8b^2, not 4b^2 as you stated above.
Either way, still fun. Math is fun, that's all I can say. 
Yes, b=1 works for that equation. The issue is that (2b^2 + 6)(2b^2 - 2) does not equal 2b^4 + 8b^2 - 12, so b=1 is not a solution to the original problem, which is what you asked. That is, again, a result of the fact that (2b^2)^2 = (2^2)(b^2)^2 = 4b^4 and not 2b^4.Switch Kicker wrote: If b = 1:
(2•1^2 + 6)(2•1^2 - 2) =>
(2•1 + 6)(2•1 - 2) =>
(2 + 6)(2 - 2) =>
8 • 0 = 0
If you want to be good at math, or any related science field that uses it (physics, chemistry, engineering etc.) you've got to be willing to make mistakes. You've got to be willing to try to solve whole problems and go back and realize that one mistake skewed your whole answer and be ok to take another stab at it. You can't let yourself get frustrated over it. Mistakes are part of the learning process, they are part of the solution process. You think someone came up with those Navier Stokes equations on the first try "gee, this is how water works". Absolutely not, they weren't even finalized until late 19th century after hundreds of years of work starting with Isaac Newton.Switch Kicker wrote:Of course. Math is crazy fun,lol.
