2x√x+16=9x. Need to find x-4/x.

Let y²=x, then 2y³+16=9y², 2y³-9y²+16=0.

When y=4, we have 128-144+16=0 so y=4 is a root. If you draw a graph of the cubic you can see that y=4 is a solution.

We can now reduce the cubic to a quadratic by dividing by y-4: 2y²-y-4.

Since y=4 and x=y², x=16 is a solution, and x-4/x=16-4/16=15¾.

The other two roots for y are (1±√33)/4.

If y=(1±√33)/4, x=(17±√33)/8 and x-4/x=±√33/4=±1.43614 approx.

The reason is that x-4/x=(17±√33)/8-32/(17±√33).

When this is rationalised: (17±√33)/8-32(17∓√33)/(289-33),

(17±√33)/8-(17∓√33)/8=±√33/4.

When we substitute x=(17±√33)/8 into the original equation we find that only (17+√33)/8 satisfies it. The other value for x doesn’t satisfy the equation because the square root is assumed to be positive. If we allow the negative square root, then the equation is satisfied. There seem to be two solutions then to the value of x-4/x: 63/4 (or 15¾ or 15.75) and √33/4=1.43614 approx.

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