Question

Consider the equation `sqrt(3t - 14) + t = 6`, are there any extraneous solutions? What is the solution for this equation?

Answer 1

Solution: `t=5`
Using standard methods, we may also find the extraneous solution `t=10`

Explanation:

Given
`color(white)("XXX")sqrt(3t-14)+t=6`

`color(white)("XXX")sqrt(3t-14)=6-t`

after squaring both sides:
`color(white)("XXX")3t-14=36-12t+t^2`

then convert into standard form
`color(white)("XXX")t^2-15+50=0`

and factor
`color(white)("XXX")(t-5)(t-10)=0`

which implies
#{: ("either",(t-5)=0," or ",(t-10)=0), (,rarr t=5,,rarr t=10) :}#

If `t=5`
then
`color(white)("XXX")sqrt(3t-14)+t=sqrt(3xx5-14)+5=sqrt(1)+5=6`
as required
`rarr t=5` is not an extraneous solution.

If `t=10`
then
`color(white)("XXX")sqrt(3t-14)+t=sqrt(3xx10-14)+10=sqrt(16)+10=14!=6`
`rarr t=10` is extraneous
The problem arises because the square root symbol (`sqrt(color(white)(x")`) is always taken to be the primary (positive) root.