Question

How do you solve `2sqrtx +1= 6/sqrtx`?

Answer 1

Reformulate as a quadratic with possible spurious solutions and solve to find `x = 9/4` (and spurious solution `x=4`)

Explanation:

Multiply both sides by `sqrt(x)` to get:

`2x+sqrt(x) = 6`

Subtract `2x` from both sides to get:

`sqrt(x) = 6-2x`

Square both sides to get:

`x = (6-2x)^2 = 36 - 24x + 4x^2`

Note that squaring may introduce spurious solutions, so need to check later.

Subtract `x` from both sides to get:

`4x^2-25x+36 = 0`

Use the quadratic formula to find:

`x = (25+-sqrt(25^2-(4xx4xx36)))/(2*4)`

`= (25+-sqrt(625 - 576))/8`

`= (25+-sqrt(49))/8`

`= (25+-7)/8`

That is `x=4` or `x=9/4`

If `x=4` then:

`2sqrt(x) + 1 = 2sqrt(4) + 1 = 4 + 1 = 5`

But `6/sqrt(x) = 6/sqrt(4) = 6/2 = 3`

So `x=4` is not a solution of the original equation.

If `x = 9/4` then:

`2sqrt(x) + 1 = 2sqrt(9/4) + 1 = 2*3/2+1 = 3 + 1 = 4`

and `6/sqrt(x) = 6/(sqrt(9/4)) = 6/(3/2) = 12/3 = 4`

So the solution of the original equation is `x = 9/4`