tttrn333 wrote:Anurag,
I just came across this problem that doesn't quite work with your method, can you explain if I'm doing it correctly?
|X^2-16|=x-4
If X^2-16>0 then X>4 or X<4 and X^2-16=x-4 Solve=> you get X=4 or X=-3. Neither fit the constraint either X>4 or X<-4
Similarly, X^2-16<0 X>-4 X<4 Solve=> X=4 or -5. Again the same issue.
Anurag@Gurome wrote:tttrn333 wrote:Anurag,
Do we use this "method" when there appears to be a infinite solution?
How does your way relate to the what we are normally taught. Case1, case2 then test values.
Anurag@Gurome wrote:tttrn333 wrote:|X-4|=4-x?
I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.
Solution:
lxl = +x if x >= 0 and = -x if x < 0
So, lx-4l = x-4 if x-4 >= 0 or x >= 4.
This implies x-4 = 4-x if x >= 4 or x = 4 is one solution.
Next, lx-4l = 4-x if x-4 < 0 or if x < 4.
So, given equation becomes 4-x = 4-x if x < 4.
This implies x < 4.
So, the solution is x <= 4.
Use this method when there is a modulus sign.
You can easily verify that the equation will not hold for x > 4.
The method definitely works.
The problem given by you is lx^2-16l = x-4.
This implies that the equation becomes
case 1) x^2 - 16 = x - 4 if x^2 - 16 >= 0 or if either
x >= 4 or x <= -4. (Note the equality sign which you have missed)
=> (x-4)(x+3) = 0 if x >= 4 or x <= -4.
Now, x = 4 or -3. So, x = 4 is a valid solution because this satisfies the inequality x >= 4.
Case 2) -x^2 + 16 = x - 4 if x^2 - 16 < 0 or if -4 < x < 4.
=> x^2 + x - 20 = 0 if -4 < x < 4
=> (x+5)(x-4) = 0 if -4 < x < 4
=> x = -5 or 4 if -4 < x < 4.
Now this is not possible. So, case 2) is rejected.
Only solution is x = 4.
