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

Jul 18, 2015

Can reformulate as a cubic equation:

$4 {x}^{3} - {x}^{2} + 16 x - 64 = 0$

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

#### Explanation:

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

$2 x \sqrt{x} = 8 - x$

Square both sides to get:

$4 {x}^{3} = {\left(8 - x\right)}^{2} = 64 - 16 x + {x}^{2}$

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

Subtract the right hand side from the left to get:

$f \left(x\right) = 4 {x}^{3} - {x}^{2} + 16 x - 64 = 0$

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

$\pm 64$, $\pm 32$, $\pm 16$, $\pm 8$, $\pm 4$, $\pm 2$, $\pm 1$, $\pm \frac{1}{2}$, $\pm \frac{1}{4}$

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 $f \left(x\right) = 0$ algebraically, but probably beyond the scope of your course.

First let ${x}_{1} = x - \frac{1}{12}$

Then:

$4 {x}_{1}^{3} = 4 {\left(x - \frac{1}{12}\right)}^{3} = 4 \left({x}^{3} - {x}^{2} / 4 + \frac{x}{48} - \frac{1}{1728}\right)$

$= 4 {x}^{3} - {x}^{2} + \frac{x}{16} - \frac{1}{432}$

So $f \left(x\right) = 4 {x}^{3} - {x}^{2} + 16 x - 64$

$= 4 {x}_{1}^{3} + \left(16 - \frac{1}{16}\right) x + \left(\frac{1}{432} - 64\right)$

$= 4 {x}_{1}^{3} + \frac{255}{16} {x}_{1} + \left(\frac{1}{432} - 64 + \frac{255}{16 \cdot 12}\right)$

This is of the form $a {x}_{1}^{3} + b {x}_{1} + c$

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

${x}_{1} = \sqrt[3]{A + \sqrt{B}} + \sqrt[3]{A - \sqrt{B}}$

hence

$x = \sqrt[3]{A + \sqrt{B}} + \sqrt[3]{A - \sqrt{B}} + \frac{1}{12}$

If you are really interested see:

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