how can you sovle the problem it's not a quadradic
2006-06-15 12:23:13 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Jeffrey D is correct. But I would like to improve his second method a little as following.
x^3-25x=504
LHS and RHS can be factored as x(x+5)(x-5)=3^2 * 2^3 * 7
this can also be written as, x(x+5)(x-5)=3^2 *14 * 2^2
Also, x(x+5)(x-5)=9 ( 9+5)( 9-5)
Comparing both sides we have, x=9
2006-06-16 01:52:26 · answer #1 · answered by shasti 3 · 0⤊ 0⤋
Let f(x) = x^3 - 25x - 504
f(x) < 0 for x=1,..., x=8 f(x) > 0 for x=10. So x is between 8 and 10. Try 9.
or
504 = 3^2 x 2^3 x 7 = x(x^2-25) = x(x-5)(x+5)
then you match the factors.
x^3 - 25x - 504 = (x-9)(x^2 + 9x + 56)
2006-06-15 13:38:42 · answer #2 · answered by Jeffrey D 2 · 0⤊ 0⤋
where (X) is you are gonna insert the 9 so it will look like this:
(9)^3-25(9)=504
that should come up with the answer which is given to you as 504.
so everytime a problem is given just like this one you just need to insert the number given in the correct letter. hopefully it helps.
2006-06-15 13:15:49 · answer #3 · answered by aaaaaaaarrrrrrrr 2 · 0⤊ 0⤋
x^3 - 25x - 504 = 0
x(x^2-25) - 504 = 0
x(x+5)(x-5)-504=0
maybe that helps? Sorry, I've kinda forgotten how to do this....
2006-06-15 12:49:27 · answer #4 · answered by Breakdancer Girl 3 · 0⤊ 0⤋
9 x 9 =81 x 9=729 ............ x^3=729
25x= 25 * x = 225
729-225=504
2006-06-15 12:28:38 · answer #5 · answered by Me.Myself.& I 2 · 0⤊ 0⤋