Question On Modulus Notation?? Mathematicians, please help!!?

Question

|2x + a| = |x| - a , a>0

Please solve the above linear equation in order to find x. Please show me the steps that u have to follow to solve it so that i can understand such questions better. Thanks a lot for ur help!

Answer 1

Rememeber |x| means either x is +ve or x is -ve.

Now on the left side of the equation, we have |2x + a| which means either 2x + a or - (2x + a).

Therefore, either 2x + a = + x - a which gives x = - 2a which cannot be as a > 0. So we discard this solution,

Or -(2x + a) = + x - a which gives - 3x = 0 or x = 0.

Hence, finally x = 0 is the solution to the given equation.

Answer 2

First of all you have given one equation in two variables
which will lead to no solution.
To say about modulus function for your studies
Then following are the points to be remembered:
1)modulus leads a positive value.
2)|x|=x ,if x>0
|x|=(-x),if x<0.
3)If you are squarring an equation in modulus problem
make sure to check the final answers by putting the
values in original equation.All values might not satisfy it.
4)Graph of modulus of any function is reflection of the
graph of original function about X-Axis.
for eg: Graph of y=|sin x| is reflection of the graph of
y=sin x about X-axis.
5)To solve such problems remove modulus sign and put
+- sign that is solve the problem taking + and then
taking - . Then check the values which satisfy the
original question.

Answer 3

1.x>0
2x+a = x-a
x=-2a(this can`t be because a is positive)

2.x<=0
2.1 x>-a/2
2x+a = -x-a
3x = -2a
x=-2a/3 >-a/2 => 2a/3 2/3<1/2(false)

2.2 x<=-a/2
-2x-a = -x-a
x=0

Answer 4

squarring the eq 4x^2+ a^2 +8*a|x| = x^2 +a^2 -2*a*|x|

Solving them would yeild 3*x^2 = -8a|x| , squarring them again
9x^4 = (8ax)^2 Solving them would give is
x = (8ax/9)^1/2

Answer 5

|2x+a| = |x| - a
or, 2x+a = +/-(|x| - a)

If, 2x+a = |x| - a
then, 2x+2a = |x|
or, +/-(2x+2a) = x......(i)

either,
x = 2x+2a
-x = 2a
x = -2a
or,
x = -2x-2a
3x = -2a
x = -(2/3)a

If, 2x+a = -|x|+a
then, -2x = |x|.......(ii)
It is cannot be possible to get any value of x, by solving (ii)

So, x = -2a, -(2/3)a