Modulus equality solving concept

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The absolute value (or modulus) | x | of a real number x is the non-negative value of x without regard to its sign.

e.g. |3| = 3 , |-3| = 3

|-2.3| = 2.3 and |8.7| = 8.7

GRAPH OF |x|,


  always remember,

|x| = x,  if x \ge 0 and

|x| = - x, if x \le 0   (why??)

take x = 2 , 2 >  0
|2| = 2, i.e. |x| = x

and, if x = -2 , -2 < 0
|-2| = -(-2) , i.e. |x| = - x
Next, if |x| = 5 , what are the solutions???
clearly there are two curve y = |x| and y = 5 and we have to find those values of x where both curve have equal values or cut each other.


therefore the equation |x| = 5 has two solutions , -5 and 5 , because both the curves y = 5 and y = |x| cut each other on -5 and 5.

therefore x = - 5 and 5
or x \in \{ - 5,5\}

Properties of Modulus:

1.|x| = | - x|

2. |xy| = |x||y|

3.\left| {\frac{x}{y}} \right| = \frac{{|x|}}{{|y|}},y \ne 0

4. {x^2} = |x{|^2}

you can check all the properties by putting any real values for x and y.

SOLUTION OF MODULUS EQUATION


|x| = x,  if x \ge 0 and

|x| = - x, if x \le 0  

we can see that for |x| , the behaviour changing point is zero, because for x \ge 0 , |x| = x and for x \le 0, |x| = - x

 means. for x \ge 0, the curve of |x| = x i.e. y = x and for x \le 0 , the curve of |x| = - x i.e. y = -x,
thus basically there are two curves or lines.

so for solving the modulus equation or inequality , first we have to find the value for which expression of mod is zero i.e. if we have |f(x)|, we have to find for what value of x , f(x) = 0 

1. |x - 1| + |x + 1| = 3

i will tell two methods for solving this equation,

 1st method,
you can use it every time,

 consider, |x - 1|
take, x - 1 = 0
therefore, x = 1
means '1' is behaviour changing point of |x - 1|,
\Rightarrow |x - 1| = - (x - 1) = 1 - x, if x \le 1
and, |x - 1| = x - 1, if x \ge 1

 now, consider
|x + 1|
take, x + 1 = 0
or, x = - 1
therefore,
|x + 1| = - (x + 1) = - x - 1, if x \le - 1
and,|x + 1| = x + 1, if x \ge - 1

 so, there are two points in this equation for which |x - 1| and |x + 1| changes their behaviour.



STEP OF SOLVE THE MODULUS EQUALITY,
here are two critical points, you may have 3 or 4
you have to take one by one beginning from left hand side as follows,
i) x \le - 1 
ii) - 1 \le x \le 1
iii)x \ge 1

because in every interval shown above one or both curve change their behaviour,

 i)x \le - 1

 equation,|x - 1| + |x + 1| = 3 becomes

 \Rightarrow 1 - x + ( - x - 1) = 3, (since, forx \le - 1, |x + 1| = - (x + 1) = - x - 1 and since x \le - 1   therefore, x is also less than 1, means  \Rightarrow |x - 1| = - (x - 1) = 1 - x)

 \Rightarrow - 2x = 3
or, x = \frac{{ - 3}}{2}

 since we consider x \le - 1, check -3/2 is less than -1 or not.
since \frac{{ - 3}}{2} < - 1
therefore, x = \frac{{ - 3}}{2} is the solution of equation

 ii)  - 1 \le x \le 1

  |x - 1| + |x + 1| = 3 becomes,
\Rightarrow 1 - x + x + 1 = 3 (since,for x \ge - 1, |x + 1| = x + 1 and |x - 1| = 1 - x as x \le 1)

 or, 2 = 3 (not possible)

 so in the interval  - 1 \le x \le 1, there is no solution of the equation,

 iii) x \ge 1

 |x - 1| + |x + 1| = 3 becomes,
x - 1 + x + 1 = 3
2x = 3
or, x = \frac{3}{2},
since \frac{3}{2} > 1 or  \frac{3}{2} \in [1,\infty ) means belongs to the interval x \ge 1
therefore x = \frac{3}{2} is the solution of the equation.

 considering i) , ii) and iii) 
solution of equation is ,x \in \{ \frac{{ - 3}}{2},\frac{3}{2}\}

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