Solve, please. Quick if you can1) − x 2 + 2 x > 12) |...

Blericker74

Blericker74

Answered

2022-07-01

Solve, please. Quick if you can
1) x 2 + 2 x > 1
2) | 1 3 x | = 5
3) | 4 x 3 | 2

Answer & Explanation

behk0

behk0

Expert

2022-07-02Added 14 answers

1) x 2 + 2 x > 1
x 2 + 2 x > 1
x 2 + 2 x 1 > 0
( x 2 2 x + 1 ) > 0
( x 2 2 x + 1 ) < 0
( x 1 ) 2 < 0

Square of a term is always positive, so there is no x R, such that (x−1)2<0.
Hence, there is no solution

2) | 1 3 x | = 5
If | x | = a and a > 0 , then x = a or x = a.
So, | 1 3 x | = 5 implies 1 3 x = 5 or 1 3 x = 5
Then, we have:
1 3 x = 5
3 x = 1 5
3 x = 4
x = 4 3
or 1 3 x = 5
3 x = 1 + 5
3 x = 6
x = 2
Therefore, the solution is x = 4 3 or x = 2
ttyme411gl

ttyme411gl

Expert

2022-07-03Added 6 answers

(3) Given: | 4 x 3 | 2

If | x | a and a > 0, then a x a
| 4 x 3 | 2 implies 2 4 x 3 2

And you get
2 4 x 3 2
2 4 x 3 and 4 x 3 2
4 x 2 + 3 and 4 x 2 + 3
4 x 1 and 4 x 5
x 1 4 and x 5 4

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