absolute value modulus from math book

Question

|x^2-4|=1

What is x?

Can someone explain step by step?

Thank you

walker · Answer

1. let's open modulus:

a) x^2-4>=0: x^2-4=1 --> x^2>=4: x^2=5 --> x = +\(sqrt{5}\), -\(sqrt{5}\). Both solutions satisfy the condition of x^2-4>=0.

b) x^2-4<0: 4-x^2=1 --> x^2<4: x^2=3 --> x = +\(sqrt{3}\), -\(sqrt{3}\). Both solutions satisfy the condition of x^2-4<0.

So, we have 4 solutions: +\(sqrt{5}\), -\(sqrt{5}\), +\(sqrt{3}\), -\(sqrt{3}\)

By the way, this link might be useful:
math-absolute-value-modulus-86462.html

ksharma12 · Answer

Yea i read through it, i just dont get the whole process of setting the equation >=0 and <0.

How did u know those were the parameters and not >0 and <=0 for example?

In addition how did u know they satisfy the condition?

walker · Answer

If there is |x|=a, to open modulus we use simple rule: if x is non-negative, x=a; if x is negative, -x=a. So, a) and b) represents non-negative and negative conditions.

In your example, let's say we consider non-negative condition x^2-4>=0 and get  sqrt{5} . To check whether the solution satisfy non-negative condition, we put it in the inequality:  (sqrt{5})^2-4>=0  --> 5-4 >=0 --> true (the solution satisfies non-negative condition).

KissGMAT · Answer

for x such it lies outside ( -2,2) solution +/- root 5

for x inside (-2,2) solution +/- root 3

absolute value modulus from math book

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