Shoutbox

quote:

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Originally posted by CookieRevised
Well, I hate to repeat stuff, but maybe this looks a bit less like 'Chinees' (no punn intended to the Chinees people ):

Given:
  x + reciprocal(x) = 30
  (where reciprocal(x) = 1/x)

Search: x

Solution:
  x + reciprocal(x) = 30
  => x + 1/x = 30
  => (x + 1/x) * x = 30 * x
  => x² + 1 = 30 * x
  => x² - 30x + 1 = 0

this is a normal quadratic equation in the form of:
  ax² + bx + c = 0     this formula is basic knowledge!
so:
  a = 1   ,   b = -30   ,   c = 1

and to find x:

     this formula is basic knowledge!

(mind the ± in front of the square root, this means there are two possible solutions)

^^ now solve that normal equation:

the discriminate (b² - 4ac):
  => b² - 4ac = (-30)² - 4 * 1 * 1
  => b² - 4ac = 896
the discriminate is positive so we will get a 'real' number. (dunno if you already studied 'complex' numbers or also called 'imaginary' numbers. But if the discriminate is negative you will take the square root out of a negative number, resulting in a complex (or imaginary) solution for x.

now to go further with calculating x:
  => x = [ 30 ± sqr(896) ] / 2
  => x = 30 / 2 ± sqr(896) / 2
  => x = 15 ± sqr(896) / sqr(4)
  => x = 15 ± sqr(896 / 4)
  => x = 15 ± sqr(224)

solution 1:  x = 15 + sqr(224)
solution 2:  x = 15 - sqr(224)


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PS: John Anderton, you made a big mistake; you forgot the square root in your formula!
"Use [-B +- (B^2 - 4 AC)^1/2] / 2A"

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oh thakns cookie, yu are so brilliant

now help me please with this problem:



(its not hard, but i would love to see how does he manage to write the solution here )