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anonymous
 4 years ago
please help!;
how do you find the solutions for x^4x^212=0
anonymous
 4 years ago
please help!; how do you find the solutions for x^4x^212=0

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(x^2  4)(x^2 + 3) = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i just want to see if im right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^2  4 = 0 (x2)(x+2) = 0 x=2,2

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.0let y = x^2 then y^2  y 12 = 0 can you solve this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait there are more solutions

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.0yup there will be 4 solution because the degree of the equation is 4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^2 + 3 = 0 x^2 = 3 \[x = i \sqrt{3}, i \sqrt{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the solutions are: \[x = 2, 2, i \sqrt{3}, i \sqrt{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0am I correct @cwrw238

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i understand why there 4 solutions but where did you get them from?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First we factored them out, didnt we? then we set each factor equal to zero

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.0It might be easier to let z = x^2 so z^2 = x^4 which allows us to go from x^4x^212=0 to z^2z12=0

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.0solve z^2z12=0 and keep in mind that z = x^2 so you can solve for x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0o because it x^2=3 its the square root plus i right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok i got it  thank you so much guys

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there you go @jim_thompson5910 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this site is really helpful: mathway.com

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.0@tegasree  u r correct
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