# Solve the equation: `sqrt(3x-4)` = 2`sqrt(x-5)` `

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### 2 Answers

You need to raise to square the equality, to remove the square root, such that:

`(sqrt(3x - 4))^2 = (2sqrt(x - 5))^2 => 3x - 4 = 4(x - 5)`

You need to open the round brackets to the right side, such that:

`3x - 4 = 4x - 4*5 => 3x - 4 = 4x - 20`

You need to move the terms that contain x to one side, such that:

`3x - 4x = 4 - 20 => -x = -16`

You need to multiplicate by -1 both sides, such that:

`x = 16`

Testing the value `x = 16` in equation yields:

`sqrt(3*16 - 4) = 2*sqrt(16 - 5)`

`sqrt(44) = 2sqrt(11)`

You may replace `4*11` for 44 such that:

`sqrt(4*11) = 2sqrt11 => 2sqrt 11 = 2sqrt 11`

**Hence, evaluating the solution to the given equation yields **`x = 16.`

`sqrt(3x -4)=2sqrt(x-5)`

subtracting both sides `sqrt(3x - 4)` we get:

`sqrt(3x-4) -sqrt( 3x-4)= 2sqrt(x-5) -sqrt(3x-4)`

`2sqrt(x-5) -sqrt(3x-4) = 0`

miltpòying boht sides by `2sqrt(x-5) + sqrt(3x -4)`

we obtain:

`(2sqrt(x-5) -sqrt(3x-4))(2sqrt(x-5) +sqrt(3x-4)) = 0`

(on the right side every number multiplyed by zero is zero)

so we have now:

`4(x-5) -(3x-4)=0`

developing:

`4x -20-3x+4 = 0`

and so on:

`x-16=0` `x=16`