Question

How do you solve `sqrt(9x+10)=x` and find any extraneous solutions?

Answer 1

`x=10` or `x=-1`
(no extraneous solutions)

Explanation:

Given
`color(white)("XXX")sqrt(9x+10)=x`

Square both sides (this where an extraneous solution might be introduced)
`color(white)("XXX")9x+10=x^2`
Rearrange as a quadratic in standard form
`color(white)("XXX")x^2-9x-10=0`
Factor
`color(white)("XXX")(x-10)(x+1)=0`
Either
`color(white)("XXX")(x-10)=0color(white)("XX")rarrcolor(white)("XX")x=10`
or
`color(white)("XXX")(x+1)=0color(white)("XX")rarrcolor(white)("XX")x=-1`

Checking against original equation:
`color(white)("XXX")`If `x=10`
`color(white)("XXXXXXX")sqrt(9x+10)=sqrt(9(10)+10)=sqrt(100)=10=x`
`color(white)("XXX")`This solution is valid.

`color(white)("XXX")`If `x=-1`
`color(white)("XXXXXX")sqrt(9x+10)=sqrt(9(-1)+10)=sqrt(1)=1=x`
`color(white)("XXX")`This solution is valid