Question

How do you solve `x-6 = (x-4)^(1/2)` and find any extraneous solutions?

Answer 1

`x=8 or 5`

If `x<4` then `x-4` is negative `=>` solution is in the complex number set of values.

Explanation:

Square both sides giving:

`(x-6)^2=x-4`

`x^2-12x+36=x-4`

`x^2-13x+40=0`

completing the square

`0=(x-13/2)^2+40 -169/4`

`0=(x-13/2)^2+160/4-169/4`

`0=(x-13/2)^2-9/4`

`x=13/2+-sqrt(9/4)`

`x=8 or 5`

Answer 2

`x=5 or x =8`

Explanation:

Remember than the index of `1/2` is the same as the square root.
We cannot find the square root of a negative number, so

`x-4 >=0 rArr x >= 4`
`color(red)("remember to check this later")`

To get rid of the square root, you need to square both sides. A common error is to square each of the terms of the left, inside of treating them as a binomial.

`(x-6)^2 = ((x-4)^(1/2))^2`

`x^2 -12x +36 = x-4 " "larr` a quadratic, so make = 0.

`x^2 -13x +40 = 0 " "larr` check for factors.

'Find factors of 40 which add (because of +40) to give 13.'
The signs will be the same, (because of +40) they are both minus (because of -13).

`"8 and 5 will work ": 8xx5=40 and 8+5=13`

`(x-5)(x-8)=0`

`"letting each factor equal 0 gives " x=5 or x =8`

`color(red)("Both answers are valid")`