Re: Some Silly Problems...

AbhyX said:

> 11. X^(1/3) - X^(1/9) = 60
> Solve for X.

In solving such problems, it is often helpful to substitute a temporary
variable which simplifies the expression.

We note that (X^(1/9))^3 is X^(1/3), and set Y = X^(1/9). Thus, we are
trying to solve: Y^3 - Y = 60.

Because we are fairly familiar with the cubes up to about 5 or so, we can
easily see that 4^3 is 64. So that tells us straight away that Y = 4.

Thus, X^(1/9) = 4.

X is therefore 4^9, or 262144.

I suggest you invest in an elementary mathematics textbook. :-)

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/2005
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
.

Relevant Pages

  • Re: how to solve this system of 3 equations?
    ... I think the form for solving c should be a quadratic, ... Get some of the c's out of the denominator by multiplying in extra ... Now substitute the value for c^3 in 2c) back into 1c), ...
    (sci.math.symbolic)
  • Re: how to solve this system of 3 equations?
    ... I think the form for solving c should be a quadratic, ... Get some of the c's out of the denominator by multiplying in extra ... Now substitute the value for c^3 in 2c) back into 1c), ...
    (sci.math.symbolic)
  • Re: Can anybody solve to make r the subject?
    ... >>method of solving the cubic. ... expand the polynomial, but when I substitute this into the general ... I have ignored the t^4 value in my rearranging as the value for t is ...
    (sci.math)
  • Re: Integral(acr sinx)^2= ?????
    ... On Fri, 9 Sep 2005, TTTomek wrote: ... > I have a difficulties in solving such an integral: ... Substitute back. ... Prev by Date: ...
    (sci.math)