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The solution set of the equation: (x2 + 3) (x2 +1) = 0 where X belongs R is......? I guess it is FI 

 Jan 12, 2019
edited by Guest  Jan 12, 2019
edited by Guest  Jan 12, 2019
 #1
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(x^2 + 3) (x^2 + 1)   = 0

 

There are no real number solutons to this......

 

Setting  both factors to 0   we have

 

x^2 + 3  = 0                    x^2 + 1 =  0

 

x^2  =  -3                       x^2 = -1

 

Note that any real number squared is either 0 or positive

 

So....no real numbers exist that make either equation true

 

 

 

cool cool cool

 Jan 12, 2019
edited by CPhill  Jan 12, 2019
 #2
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Thank you !

Guest Jan 12, 2019

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