Originally Posted by

**JiaJian** I have a question regarding the steps to solve this |x| + |x+2| = 3 (ans x= -2.5, 0.5)

I did the first method which was to use the concept of |x| = a, therefore x = a or x = -a.

but the second method was to consider 3 regions:

x < -2, -2 <= x < 0, x >= 0

and that was what I don't really understand. My teacher said it was the JC way of doing it, can somebody explain it to me?

Thanks!

If this is the AM questions, den i try using the Amath method which should be much longer but within ur knowledge.

Knowing that to remove the modulus, one way is to square it. I.e. |x| = a => x^2 = a^2.

we have.

|x| + |x+2| = 3

By squaring both side,

x^2 + 2|x||x+2| + (x+2)^2 = 9

2|x||x+2| = 9 - x^2 - (x+2)^2

2|x||x+2| = 9 - x^2 - x^2 - 4x - 4

2|x||x+2| = -(2x^2 + 4x +5)

By squaring both side again,

4(x^2)(x+2)^2 = (2x^2 + 4x +5)^2

well, expand both side, simplify it and you may be able to solve.

Above are the standard method of solving the stated questions for Amath. As you can see JC methods are simpler.