Question

How do you solve `sqrt(3x)+8=x+2`?

Answer 1

`x={3,12}`

Explanation:

`sqrt(3x)+8=x+2`
`sqrt(3x)=x+2-8`
`sqrt(3x)=x-6`
`(sqrt(3x))^2=(x-6)^2`
`3x=x^2-12x+36`
`x^2-12x-3x+36=0`
`x^2-15x+36=0`
`(x-12)(x-3)=0`
`"if (x-12)=0 then x=12"`
`"if (x-3)=0 then x=3"`
`x={3,12}`

Answer 2

3 and 12

Explanation:

`sqrt(3x) + 8 = x + 2`
Isolate the radical term.
`sqrt(3x) = x - 6`
Square both sides:
`3x = (x - 6)^2 = x^2 - 12x + 36`
`x^2 - 15x + 36 = 0`
Find 2 numbers (real roots) knowing sum (15 = -b) and product (c = 36). They are: 3 and 12.