I dont know how to start this equation

Question

√3x) + 6 = x

The 3x is under the radical sign.

Answer 1

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.

 

Answer 2

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

 

 

Answer 3

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).