Question

How do you solve `x = sqrt(-3x + 10)` and find any extraneous solutions?

Answer 1

`x=-5" and "x=+2" "` are solutions to `x^2+3x-10=0`

Extraneous `ul("values")` of `x>10/3`

For `AAx notin {-5,2}" " =>" " x^2+3x-10 != 0`

Explanation:

`color(blue)("Determine general solution")`

Square both sides

`x^2=-3x+10" "larr" you now have a quadratic"`

`x^2+3x-10=0`

Notice that `2xx5=10" and "5-2=+3`

Thus the factorization is`" "(x+5)(x-2)=0`

So `x=-5" and "x=+2` are general solutions
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine extraneous values of "x)`

For the number to remain in the `ul("'Real'")` domain you may can not permit the square root to be that of a negative number.

So `-3x+10>=0`

Multiply by (-1) giving

`+3x-10<=0`
Notice that the inequality sign has turned the other way. This happens when you multiply by (-1)

Add 10 to both sides

`3x<=10`

`=>x<=10/3` as a required condition

`=>x>10/3` takes the solution out of the Real domain and into the Complex number domain