Question

How do you solve `sqrt(2X-1) = 5 - sqrt(X-1)` and find any extraneous solutions?

Answer 1

`color(green)(x=5)`
`color(white)("XXX")`extraneous solution at `x=145`

Explanation:

Given
`color(white)("XXX")sqrt(2x-1)+sqrt(x-1)=5`

Squaring both sides
`color(white)("XXX")2x-1+2sqrt(2x-1)sqrt(x-1)+x-1=25`

Simplifying
`color(white)("XXX")3x-2+2sqrt(2x^2-3x+1)=25`

`color(white)("XXX")2sqrt(2x^2-3x+1)=27-3x`

Squaring both sides again
`color(white)("XXX")8x^2-12x+4=729-162x+x^2`

Simplifying into standard quadratic form
`color(white)("XXX")x^2-150x+725`

Factoring (or use the quadratic formula if the factors are not apparent)
`color(white)("XXX")(x-5)(x-145)=0`

Giving possible solutions:
`color(white)("XXX")x=5color(white)("XX")andcolor(white)("XX")x=145`

Substituting back into the original equation:
[1]`color(white)("XXX")x=5`
`color(white)("XXXXX")LS`
`color(white)("XXXXXXX")=sqrt(2*5-1)`
`color(white)("XXXXXXX")=sqrt(9)`
`color(white)("XXXXXXX")=3`
`color(white)("XXXXX")RS`
`color(white)("XXXXXXX")=5-sqrt(5-1)`
`color(white)("XXXXXXX")=5-2`
`color(white)("XXXXXXX")=3`
`color(white)("XXX")LS=RScolor(white)("XXX")`valid

[2]`color(white)("XXX")x=145`
`color(white)("XXX")LS`
`color(white)("XXXXX")=sqrt(2*145-1)`
`color(white)("XXXXXXX")=sqrt(289)`
`color(white)("XXXXXXX")=17`
`color(white)("XXX")RS`
`color(white)("XXXXXXX")=sqrt(145-1)`
`color(white)("XXXXXXX")=sqrt(144)`
`color(white)("XXXXXXX")=12`
`color(white)("XXX")LS!=RScolor(white)("XXX")`extraneous