I see..3/|2| > 1
Does it follow that 3 > 2^2?
great input once again steven g!Squaring just one side of an inequality usually doesn't work.
You are given:Ought this not be allowed, I realise now that I didn't have to times both sides by the modulus of that squared since you know it's always going to be positive anyways but does the definition of a modulus not work in practiceView attachment 37570
Squaring an inequality troubles me; this will result in extraneous solutions.You are given:
(5x-1)/|(2x-3)| >= 1
(5x-1)^{2}/(2x-3)^{2} >= 1^{2} >= 1 ......................... "modulus" sign removed
(5x-1)^{2} >= (2x-3)^{2}
25*x^{2} - 10*x + 1 >= 4*x^{2} - 12*x + 1
21*x^{2} + 2*x >= 0
x * (21*x + 2) >= 0
Now investigate the equation above for the set values of x that would satisfy the given condition. Do not forget to check .................
There’s an error in mine as well but wait so then can I actually treat the modulus as just the square root of a squared term? even if it is more complicated?Squaring an inequality troubles me; this will result in extraneous solutions.
There's also an arithmetic error.
My preferred approach is to multiply both sides by the (positive) denominator, and then take it in two cases, according to whether 2x-3>0 or 2x-3<0.
Neither of us is recommending using square roots; we're doing very different things from that. The closest we come is in saying that if you square both sides, the square of |2x-3| is (2x-3)^2, because squaring loses information about the sign.so then can I actually treat the modulus as just the square root of a squared term? even if it is more complicated?
Sorry abut that .....arithmetic error.