Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# What are the restrictions to √3x−1 + 5 = 2?

The 3x-1 are under the square root sign

Relevance
• 2 months ago

3x - 1 >= 0

x >= 1/3

for this particular equation,

√(3x−1) + 5 = 2

√(3x - 1) = -3... no solution

• ted s
Lv 7
2 months ago

x ≥ 1 / 3...but there are no answers as 5 > 2

• ?
Lv 7
2 months ago

If (3x-1) is under the root sign

then

3x-1 ≥ 0

3x ≥ 1

x ≥ 1/3

• A.J.
Lv 7
2 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.