Question

How to solve this? `sqrt(x-2)+root(3)(4-x)=2`

Answer 1

`x = 3" "` or `" "x = 14+-6sqrt(3)`

Explanation:

Transform from a radical equation to a polynomial in several steps, find the solutions, then check they are not extraneous...

Given:

`sqrt(x-2)+root(3)(4-x) = 2`

Subtract `sqrt(x-2)` from boh sides to get:

`root(3)(4-x) = 2-sqrt(x-2)`

Cube both sides to get:

`4-x = (2-sqrt(x-2))^3`

`color(white)(4-x) = 8-12sqrt(x-2)+6(x-2)-(x-2)sqrt(x-2)`

`color(white)(4-x) = 6x-4-(x+10)sqrt(x-2)`

Add `(x+10)sqrt(x-2)+x-4` to both sides to get:

`(x+10)sqrt(x-2)=7x-8`

square both sides to get:

`(x^2+20x+100)(x-2) = 49x^2-112x+64`

Multiplying out, this becomes:

`x^3+18x^2+60x-200 = 49x^2-112x+64`

Subtract the right hand side from the left to get:

`0 = x^3-31x^2+172x-264`

`color(white)(0) = (x-3)(x^2-28x+88)`

`color(white)(0) = (x-3)(x^2-28x+196-108)`

`color(white)(0) = (x-3)((x-14)^2-(6sqrt(3))^2)`

`color(white)(0) = (x-3)((x-14)-6sqrt(3))((x-14)+6sqrt(3))`

`color(white)(0) = (x-3)(x-14-6sqrt(3))(x-14+6sqrt(3))`

So:

`x = 3`

or:

`x = 14+-6sqrt(3)`

Trying these values in the original equation, we find that `x=3` is a valid solution.

How about the other two?

Note that cubing the equation would introduce no extraneous solutions since `f(x) = x^3` is one to one. Squaring the equation introduces extraneous solutions if and only if `(7x-8)/(x+10) < 0`

We find:

`7(color(blue)(14+6sqrt(3)))-8 = 90+42sqrt(3) > 0`

`color(blue)(14+6sqrt(3))+10 = 24+6sqrt(3) > 0`

So `14+6sqrt(3)` is a valid solution.

`7(color(blue)(14-6sqrt(3)))-8 = 90-42sqrt(3) > 0`

`color(blue)(14-6sqrt(3))+10 = 24-6sqrt(3) > 0`

So `14-6sqrt(3)` is a valid solution too.

graph{(y-sqrt(x-2)-root(3)(4-x))(y-2) = 0 [-1, 33.74, -1, 2.5]}