What are the restrictions to √3x−1 + 5 = 2?
The 3x-1 are under the square root sign
- Iggy RockoLv 72 months ago
3x - 1 >= 0
x >= 1/3
for this particular equation,
√(3x−1) + 5 = 2
√(3x - 1) = -3... no solution
- ted sLv 72 months ago
x ≥ 1 / 3...but there are no answers as 5 > 2
- ?Lv 72 months ago
If (3x-1) is under the root sign
3x-1 ≥ 0
3x ≥ 1
x ≥ 1/3
- A.J.Lv 72 months ago
Since it is the square root of (3x-1), 3x-1 cannot be less than zero or it becomes an imaginary number as a square root. 3x-1=0 >>> x = 1/3, and as x gets larger 3x-1 gets larger, so x >= 1/3
Your other issue is that if sqrt (3x-1) +5 = 2, then subtracting 5 from both sides says sqrt (3x-1) = -3
Squaring both sides says 3x-1 = 9 and 3x=10 so x = 10/3
but (sqrt 9) +5 = 8
I don't remember what this is called when only the negative applies and how to explain it, but in this equation it has to at least be noted.
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- Donnie PorkoLv 72 months ago
X cannot be less than 0 because then you get square root of a negative number.