promotion image of download ymail app
Promoted
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

5 Answers

Relevance
  • 2 months ago

    3x - 1 >= 0

    x >= 1/3

    for this particular equation,

    √(3x−1) + 5 = 2

    √(3x - 1) = -3... no solution

    • Commenter avatarLogin to reply the answers
  • ted s
    Lv 7
    2 months ago

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

    • Commenter avatarLogin to reply the answers
  • ?
    Lv 7
    2 months ago

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

    then

    3x-1 ≥ 0

    3x ≥ 1

    x ≥ 1/3

    answer

    • Commenter avatarLogin to reply the answers
  • 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.

     

    • Commenter avatarLogin to reply the answers
  • How do you think about the answers? You can sign in to vote the answer.
  • 2 months ago

    X cannot be less than 0 because then you get square root of a negative number. 

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.