by 
Masta
ANSWERS: 6
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Since you are taking the absolute value of (x^4)-1, either (x^4)-1 or -((x^4)-1) will work, so that will give you two different inequalities. Perform the same operation on the left and right sides of the inequality to simplify it until you have the solution. If this is a homework problem you have to do it yourself.
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x^4 - 1 is a difference of squares. A difference of squares a^2-b^2 can be factorized: (a-b)*(a+b) x^4 - 1 is (x^2-1)*(x^2+1) Let C stand for x^2+1, which is always positive. Now we have: abs((C-2)*C) <= 3C and since C is always positive: abs((C-2))*C <= 3C and also abs(C-2) <= 3 if the distance of C from 2 is less than or equal to 3 then C can be anywhere from -1 to 5 (inclusive) if x^2 +1 is anywhere from -1 to 5 (inclusive) then x^2 is anywhere from 0 to 4 (inclusive) and x is anywhere from -2 to 2 (inclusive) EDIT: You reached the conclusion: (x^4)<=3(x^2)+4 You can always add and subtract from inequalities: x^4 - 3 x^2 - 4 <= 0 The left side can be made into a square. If it were x^4 - 3 x^2 + (9/4) it would be (x^2 - (3/2))^2 So... x^4 - 3 x^4 <= 4 x^4 - 3 x^4 + (9/4) <= 4 + (9/4) (x^2 - (3/2))^2 <= 25/4 abs(x^2 - 3/2) <= 5/2 -5/2 <= x^2 - 3/2 <= 5/2 -1 <= x^2 <= 4 x^2 <= 4
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Im not sure all solutions are solved in the first answer. When a function, f(x) is enclosed in abs, there are 2 solutions. For example: |f(x)|>=C There are 2 solutions for this. In the case that f(x) is positive, |f(x)|>=C in the case that f(x) is negative, |f(x)|<=C [This is because when multiplying an inequality by a negative, the inequality must switch]. Note there maybe more than wone regions of positive and negative. In this case you must specify domains. (i.e. positive when 0<x<3, negative elsewhere). Edit: I just checked Quirkie's answer. He seems to have gotten both solutions regardless. I'll leave this answer as it may help others with inequality problems.
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I honestly tried... and my brain just exploded. Good luck.
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Pay attention in class instead of passing notes or looking at the teachers breast??....that might help solve the equation...;)
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abs(x^4-1)<=3(x^2+1) let x=Ai +B (A and B real numbers, i basis of imaginary numbers) x^2= -A^2+2ABi+B^2 if A<>0 and B<>0 then ABi is imaginary. As there is no order relation on the set of complex number, the inequality must be false. So the inequality can only be true if A=0 or B=0 1) B=0 (or: x = A) abs(A^4-1) <= 3(A^2+1) abs((A^2-1)(A^2+1)) <= 3(A^2+1) abs(A^2-1) <= 3 -3 <= A^2-1 <= 3 -2 <= A^2 <= 4 -2 <= A <= 2 -2 <= x <= 2 2) A=0 (or: x = iB) abs(B^4-1)<=3(-B^2+1) abs((B^2-1)(B^2+1)) <= 3(-B^2+1) 1-B^2 = (1-B)(1+B) a) B > 1 yields 1-B^2 < 0 no solutions! b) B >= 1 yields 1-B^2 > 0 - 3(1-B^2) <= (1-B^2)(B^2+1) <= 3(1-B^2) -3 <= B^2+1 <= 3 -2 <= B <= 2 only possible if x = 0 (no order relation by complex numbers!)
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