Hmmmm...I'm not sure I follow.
This is how I solved the problem.
|x^2-4|=1
x^2 - 4 =1
OR
-x^2 + 4 = 1
SO
x^2=5 ==> x=+/- \sqrt{5}
OR
-x^2=-3 ==> x^2=3 ==> x=+/- \sqrt{3}
So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!
This is how I solved the problem.
|x^2-4|=1
x^2 - 4 =1
OR
-x^2 + 4 = 1
SO
x^2=5 ==> x=+/- \sqrt{5}
OR
-x^2=-3 ==> x^2=3 ==> x=+/- \sqrt{3}
So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!
Zarrolou wrote:
The first method is the correct one and will always give you the correct results.
Consider however the following case
|x+5|=-4, at glance this equation has no solution because |x+5| cannot be less than 0.
But I wanna take it as example:
With the first method you'll find
if x>-5
x+5=-4, x=-9, out of the interval => it's not a solution
if x<-5
-x-5=-4, x=-1 out of the interval => it's not a solution
With the second method
x+5=-4, x=-9
-(x+5)=-4, or x=-1
those seem valid... but the equation we know that has no solution.
Main point: the first method works always, do not rely on the other one.
The second one does not take into consideration the intervals, so it might not work
Hope it's clear
Consider however the following case
|x+5|=-4, at glance this equation has no solution because |x+5| cannot be less than 0.
But I wanna take it as example:
With the first method you'll find
if x>-5
x+5=-4, x=-9, out of the interval => it's not a solution
if x<-5
-x-5=-4, x=-1 out of the interval => it's not a solution
With the second method
x+5=-4, x=-9
-(x+5)=-4, or x=-1
those seem valid... but the equation we know that has no solution.
Main point: the first method works always, do not rely on the other one.
The second one does not take into consideration the intervals, so it might not work
Hope it's clear