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tejasreeBest ResponseYou've already chosen the best response.1
(x^2  4)(x^2 + 3) = 0
 one year ago

dddanBest ResponseYou've already chosen the best response.0
i just want to see if im right
 one year ago

tejasreeBest ResponseYou've already chosen the best response.1
x^2  4 = 0 (x2)(x+2) = 0 x=2,2
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
let y = x^2 then y^2  y 12 = 0 can you solve this?
 one year ago

tejasreeBest ResponseYou've already chosen the best response.1
wait there are more solutions
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
yup there will be 4 solution because the degree of the equation is 4
 one year ago

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

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

tejasreeBest ResponseYou've already chosen the best response.1
am I correct @cwrw238
 one year ago

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

tejasreeBest ResponseYou've already chosen the best response.1
First we factored them out, didnt we? then we set each factor equal to zero
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
It 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
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
solve z^2z12=0 and keep in mind that z = x^2 so you can solve for x
 one year ago

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

dddanBest ResponseYou've already chosen the best response.0
ok i got it  thank you so much guys
 one year ago

tejasreeBest ResponseYou've already chosen the best response.1
there you go @jim_thompson5910 :)
 one year ago

tejasreeBest ResponseYou've already chosen the best response.1
this site is really helpful: mathway.com
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
@tegasree  u r correct
 one year ago
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