# If root(3)(3(root(3)x - 1/(root(3)x))) = 2, then root(3)x + 1/(root(3)x) = what?

Feb 4, 2015

$\sqrt{3 \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)} = 2$

and we want to solve for:

$\sqrt{x} + \frac{1}{\sqrt{x}}$ (note the + instead of -)

which means we need to do a bit more work and solve for $x$ and then use our x value to solve for:

$\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)$

We then take each side to the exponent 3 (which gets rid of the cube root on the left hand side):
${\left(\sqrt{3 \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)}\right)}^{3} = {2}^{3}$

Taking the cube root of a number is equivalent to taking a fractional exponent of 1/3 and so this simplifies to:
${\left(3 \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)\right)}^{\frac{3}{3}} = 8$

$3 \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right) = 8$

We then divide both sides by 3 (to get ride of the multiplication by 3 on the left hand side)
$3 \frac{\sqrt{x} - \frac{1}{\sqrt{x}}}{3} = \frac{8}{3}$

$\sqrt{x} - \frac{1}{\sqrt{x}} = \frac{8}{3}$

Next we multiply both sides by $\sqrt{x}$ to remove it from the denominator of the second term

${\sqrt{x}}^{2} - 1 = \frac{8 \cdot \sqrt{x}}{3}$

${\sqrt{x}}^{2} - \frac{8 \cdot \sqrt{x}}{3} - 1 = 0$

You might not immediately recognize it, but this is a quadratic equation. To make that more obvious, we're going to temporarily substitute $\sqrt{x}$. Let's choose a new variable $y$ to represent $\sqrt{x}$:

$y = \sqrt{x}$

And now our equation becomes:

${y}^{2} - 8 \frac{y}{3} - 1 = 0$

which more closely resembled what we expect from a quadratic equation. Using the quadratic formula and the letter $y$ now instead of the traditional $x$:

 y = (-b ±sqrt(b^2 - 4ac))/(2a)

we find that:

$y = 3$ and $y = - \frac{1}{3}$

Now we'll substitute back our function of x for y: $y = \sqrt{x}$

$\sqrt{x} = 3$
$x = {3}^{3}$
$x = 27$

and

$\sqrt{x} = - \frac{1}{3}$
$x = {\left(- \frac{1}{3}\right)}^{3}$
$x = - \frac{1}{27}$

So we have two solutions for x:
$x = 27$ and $x = - \frac{1}{27}$

Using either of these values ($27 \mathmr{and} - \frac{1}{27}$), we should be able to find the answer to the original problem:

$\sqrt{x} + \frac{1}{\sqrt{x}}$
$= \sqrt{27} + \frac{1}{\sqrt{27}}$

We know that $\sqrt{27} = 3$ which means:

$= 3 + \frac{1}{3}$

$= \frac{9 + 1}{3}$

$= \frac{10}{3}$