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Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0I know the square root argument must be grater to 0 but I can solve the inside for x> 0 .

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0@Callisto @UnkleRhaukus

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x^4+3x^2+4}+1\]

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0That's the right one sorry :P .

Aylin
 2 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt{x^{4}+3x^{2}+4}\] \[x^{4}+3x^{2}+4 \ge 0\]I'm going to try to divide the equation by (x^2+1)\[\frac{ x^{4}+3x^{2}+4 }{ x^{2}+1 }=(x^{2}4)=(x+2)(x2)\](The above was done on paper since it would be too tedious to put in the calculation here). But this shows that \[x^{4}+3x^{2}+4=(x+2)(x2)(x^{2}+1) \ge 0\]You should be able to get the rest. (x^2+1 was attempted since all of the powers of x were divisible by 2, and all of the coefficients had 1 as their only common factor).
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