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

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tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1(x^2  4)(x^2 + 3) = 0

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

tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1x^2  4 = 0 (x2)(x+2) = 0 x=2,2

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

tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1wait there are more solutions

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

tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1x^2 + 3 = 0 x^2 = 3 \[x = i \sqrt{3}, i \sqrt{3}\]

tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1the solutions are: \[x = 2, 2, i \sqrt{3}, i \sqrt{3}\]

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

tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1First we factored them out, didnt we? then we set each factor equal to zero

jim_thompson5910
 3 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
 3 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

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

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

tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1there you go @jim_thompson5910 :)

tejasree
 3 years ago
Best ResponseYou've already chosen the best response.1this site is really helpful: mathway.com

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