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1/13/2013 6:42:59 PM
4
Need a little math help.
x^4-8^2+16=0
I know the answer is +-2, but how do I get to it?
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Well, from observation, I know that -4 * -4 = 16, and that -4 + -4 = -8, so I know that this will factor into (x^2 - 2)(x^2 - 2), and each of those factor into (x+2)^2(x-2)^2. Whenever you have some form of Ax^2 + Bx + C, you always want to ask yourself first: "Are there two numbers that add together to form B that multiply together to form C?"
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Edited by Pulse Cloud: 1/13/2013 7:03:07 PMYou're wrong, OP. (-2)^4 -8^2 + 16 = 2^4 - 8^2 + 16 = -32 1. -8^2 + 16 = -48 2. x^4 - 48 = 0 <=> x^4 = 48 3. x^2 = sqrt(48) = sqrt(16 * 3) = sqrt(4^2 * 3) = 4 * sqrt(3) 4. x = +- sqrt(4 * sqrt(3)) = +- sqrt(4) * sqrt(sqrt(3)) = +- 2 * 3^(1/4) [quad root] Real solutions: +- 2 * 3^(1/4) Complex solutions: They exist, but I don't remember how to calculate them... Sorry.
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Cant tell from the way its written is it to the fourth power and is the 8 squared. Its confusing
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[url=http://www.wolframalpha.com/input/?i=x%5E4-8%5E2%2B16%3D0]Here.[/url]
