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I dont know how to start this equation

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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

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

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).