Question

How do you solve `\sqrt { 54+ 3x } = x`?

Answer 1

`x=9 or x=-6`

Explanation:

Square both sides

`54+3x=x^2`

Take `54` and `3x` from both sides

`x^2-3x-54=0`

Then we Differentiate

`(x-9)(x+6)=0`

`x=9 or x=-6`

Answer 2

`x_1=9`; `x_2=-6`

Explanation:

Squaring both sides, we can get the left side of the equation out of the square root.
`54+3x=x^2`
`x^2-3x-54=0`
`(x-9)(x+6)=0`

Hence, the roots of `x` are:
`x_1=9` and `x_2=-6`

Answer 3

`color(red)(x=9 and x=-6`

Explanation:

`sqrt(54+3x)=x`

`:.(sqrt(54+3x))^2=x^2`

`:.sqrt(54+3x)*sqrt(54+3x)=x^2`

Note:

`color(red)(sqrta*sqrta=a`

`:.x^2=54+3x`

`:.x^2-3x-54=0`

`:.(x-9)(x+6)=0`

`:.color(red)(x=9 and x=-6`