Question

How do you solve `6 - sqrt (3x) = -3`?

Answer 1

`x=27`

Explanation:

Given
`color(white)("XXX")6-sqrt(3x)=-3`

Adding `(sqrt(3x)+3)` to both sides (and exchanging the sides)
`color(white)("XXX")sqrt(3x)=9`

Squaring both sides:
`color(white)("XXX")3x=81`

Dividing both sides by `3`
`color(white)("XXX")x=27`

Answer 2

`x=27`

Explanation:

First, you isolate the root in one side of the equation by subtracting 6 to both sides:

`6-sqrt(3x)=-3`

`-sqrt(3x)=-3-6=-9`

You can eliminate the minus sines by multiplicating both side by -1:

`sqrt(3x)=9`

The next step is to square both sides of the equation. This can add false solutions to the equation, so be aware that the dominion is x>0.

`(sqrt(3x))^2=9^2`

`3x=81`

divide both sides by 3, you will get the answer:

`x=27` (is larger than zero, so is a valid solution)

Answer 3

You must isolate the radical to one side of the equation before squaring to get rid of it.

Explanation:

It would probably be easier to send the radical to the other side and the 3 to be with the 6, because it is simpler to work with a positive radical.

`6 + 3 = sqrt(3x)`

`9 = sqrt(3x)`

Now we square to eliminate the square root:

`(9)^2 = (sqrt(3x))^2`

`81 = 3x`

`27 = x`

So, x = 27.

N.B: Always check your solutions in the original equation. When dealing with more complex questions, usually the type where there are two solutions, oftentimes only one or neither solution will work. Solutions that don't work in the original equation are called extraneous solutions.

Practice exercises:

1. Solve for x. Beware of extraneous solutions.

a) `sqrt(2x + 3) = 5`

b) `sqrt(3x + 1) = x - 1`

c) `sqrt(3x) - sqrt(2x + 1) = 1`

d) `sqrt(12x + 1) + sqrt(17x - 4) = 15`

Good luck!