baboosaa
I was learning my maths lesson. In an entrance preparation book i saw this question:
x< x*x< 2x + 1
x< x*x< 2x + 1

quadratic inequailtiesbaboosaa
I was learning my maths lesson. In an entrance preparation book i saw this question:
x< x*x< 2x + 1 Bikerman
0<x<1+root2
infinisa
Right answer, but I suspect baboosaa would like to know how one gets the answer: First, we separate into 2 inequalities, both of which must be satisfied: a) x< x^2 b) x^2< 2x + 1 Starting with a), we have: x^2 + x > 0 which factorises to x(x + 1) > 0 The graph of the left side is a a parabola, with +ve values outside the zeros (Algebraically, between the 2 zeros, the terms have opposite sign and so negative product) To find the zeros, use factorised form: x (x + 1) = 0 The zeros are x= 0 and x= 1 So a) is true if x < 1 OR x > 0 Similarly for b), we have: x^2  2x  1 < 0 Once again, the graph of the left side is a a parabola, with ve values between the zeros This time, we can't easily factorize, so must use the quadratic formula to find the zeros, which gives: (2 ± √(8))/2 which simplifies to 1 ± √(2) (as √(8) = 2√(2)) So b) is true if 1  √(2) < x < 1 + √(2) Since a) and b) must both be true, and 1 < 1√(2), we get 0 < x < 1 + √(2) as per Bikerman Hope this helps Bikerman
I thought providing the answer in this case might stimulate the questioner to play around with it to see where it came from
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