|x^2 - 2| < 1 Please can anyone help? How to solve this absolute value problem?

2 Answers
Narad T.
Mar 15, 2018

The solution is x in (-sqrt3,-1)uu(1, sqrt3)x∈(−√3,−1)∪(1,√3)

Explanation:

|x^2-2|<1∣∣x2−2∣∣<1

There are 22 solutions

x^2-2<1x2−2<1 and x^2-2> -1x2−2>−1

x^2-3<0x2−3<0,=>⇒ x in (-sqrt3, sqrt3)x∈(−√3,√3)

x^2-1>0x2−1>0, =>⇒, x in (-oo,-1)uu(1,+oo)x∈(−∞,−1)∪(1,+∞)

You can prove theses solutions by a sign chart.

Combining the 22 solutions

x in (-sqrt3,-1)uu(1, sqrt3)x∈(−√3,−1)∪(1,√3)

graph{|x^2-2|-1 [-6.24, 6.25, -3.117, 3.126]}

Cesareo R.
Mar 15, 2018

See below.

Explanation:

Squaring both sides

(x^2-2)^2 < 1 rArr (x^2-2)^2 - 1 < 0 rArr (x^2-2-1)(x^2-2+1) < 0(x2−2)2<1⇒(x2−2)2−1<0⇒(x2−2−1)(x2−2+1)<0 or

(x^2-3)(x^2-1) < 0(x2−3)(x2−1)<0 which is equivalent to

{(x^2-3 < 0),(x^2-1>0):} rArr 1 < x^2 < 3

the set

{(x^2-3 > 0),(x^2-1<0):} rArr {x^2 < 1} nn {x^2 > 3}

is an extraneous solution, is not feasible and was included due to the squaring operation.

Finally

1 < x^2 < 3

Related questions
Impact of this question
1310 views around the world
You can reuse this answer
Creative Commons License