Question

How do you solve `2(xsqrt x) + x=8`?

Answer 1

Can reformulate as a cubic equation:

`4x^3-x^2+16x-64 = 0`

which has an irrational root findable using Cardano's method or similar.

Explanation:

Subtract `x` from both sides of the equation to get:

`2xsqrt(x) = 8-x`

Square both sides to get:

`4x^3 = (8-x)^2 = 64 - 16x+x^2`

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

Subtract the right hand side from the left to get:

`f(x) = 4x^3-x^2+16x-64 = 0`

By the rational roots theorem, the only possible rational roots of this cubic are:

`+-64`, `+-32`, `+-16`, `+-8`, `+-4`, `+-2`, `+-1`, `+-1/2`, `+-1/4`

None of these are roots (though `2` comes quite close).

graph{4x^3-x^2+16x-64 [-0.698, 4.302, -1.18, 1.32]}

It is possible to solve `f(x) = 0` algebraically, but probably beyond the scope of your course.

First let `x_1 = x - 1/12`

Then:

`4x_1^3 = 4(x-1/12)^3 = 4(x^3-x^2/4+x/48-1/1728)`

`=4x^3-x^2+x/16-1/432`

So `f(x) = 4x^3-x^2+16x-64`

`= 4x_1^3 + (16-1/16)x+(1/432-64)`

`= 4x_1^3 + 255/16x_1+(1/432-64+255/(16*12))`

This is of the form `ax_1^3+bx_1+c`

Then you can use Cardano's method to solve, finding a solution of the form:

`x_1 = root(3)(A+sqrt(B)) + root(3)(A-sqrt(B))`

hence

`x = root(3)(A+sqrt(B)) + root(3)(A-sqrt(B)) + 1/12`

If you are really interested see:

https://en.wikipedia.org/wiki/Cubic_function