They want x-4/x
asked 4 days ago in Word Problem Answers by Zen

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

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),


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.

answered 4 days ago by Rod Top Rated User (559,680 points)

Related questions

Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
81,275 questions
85,406 answers
68,895 users