Question

How do you solve `2x - 9\sqrt { x } + 4= 0`?

Answer 1

See below.

Explanation:

Move the term with the radical to the right hand side:

`2x + 4 = 9sqrt(x)`
square both sides and collect like terms:
`81x = 4x^2 + 16x +16`
collect and equate to 0:
`4x^2 -65x + 16 = 0`
Factor:
`(4x - 1)(x -16) = 0`
solution: `x = 16 , x = 1/4`

Answer 2

`x=16" "` or `" "x = 1/4`

Explanation:

Given:

`2x-9sqrt(x)+4 = 0`

Let `t=sqrt(x)` and find:

`0 = 8(2x-9sqrt(x)+4)`

`color(white)(0) = 8(2t^2-9t+4)`

`color(white)(0) = 16t^2-72t+32`

`color(white)(0) = (4t)^2-2(4t)(9)+9^2-49`

`color(white)(0) = (4t-9)^2-7^2`

`color(white)(0) = ((4t-9)-7)((4t-9)+7)`

`color(white)(0) = (4t-16)(4t-2)`

`color(white)(0) = 8(t-4)(2t-1)`

So:

`t=4" "` or `" "t=1/2`

Then:

`x = t^2 = 4^2 = 16" "` or `" "x = t^2 = (1/2)^2 = 1/4`