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I dont know how to start this equation

√3x) + 6 = x

The 3x is under the radical sign.

Comments

Please confirm that you mean that only the 3 is under the square root sign, not the x. If that is the case, Shawn's method seems like the most straight forward.

Your problem is tagged "Completing the Square", which makes me wonder if there isn't a missing exponent.

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3 Answers

Hello Lucinda,

I'm guessing 3x is under the radical sign. Here is the solution step by step.

  • Subtract 6 from both sides            √3x + 6 - 6 = x - 6

√3x = x - 6

  • Square both sides                        (√3x)2 = (x - 6)2

((3x)1/2)2 = (x - 6)2 (converted radical into exponential form)

(3x)2/2 = (x - 6)2

3x = x2 - 12x + 36        ( (a - b)2 = a2 - 2ab + b2)

  • Next subtract 3x from both sides            3x - 3x = x2 - 12x -3x + 36

0 = x2 -12x -3x + 36

or x2 -12x -3x + 36 = 0

  • Factorize it                                         x(x - 12) -3(x - 12) = 0
  • Factor by grouping                              (x - 12)(x - 3) = 0 (answer)

I hope this helps.

 

Comments

Comment

Shawn may have misinterpretted the first term. It is sqrt(3X), not sqrt(3) x X. Therefore,

sqrt(3X) + 6 = X               now subtract 6 from each side

sqrt(3X) = X - 6                now square both sides

3X = X^2 - 12X + 36        subtract 3X from both sides

0 =   X^2 - 15X + 36         or

X^2 - 15X + 36 = 0          now factor or use quadratic formula. Factoring yields

(X - 3)(X- 12) = 0            now set each factor = 0

(X - 3) = 0 and X = 3 or (X - 12) = 0 and X = 12

 

 

Comments

Comment

You move terms w/ variables to one side and constant term to the other side:

   sqrt(3) x -x = -6

Factor out x:

   (sqrt(3)-1) x = 6

divide the (sqrt(3)-1) from both sides:

   x = 6 / (sqrt(3)-1)

multiply sqrt(3)+1 to both numerator and denominator:

  x = 3 (sqrt(3)+1).