x^3 - 729 = 0
2007-11-24 11:01:43 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It asks for more then just 9, there is also an equation with something over something and the top has an i in it and there is a square root sign I don't know how to get there.
2007-11-24 11:07:52 · update #1
It is something over 2
2007-11-24 11:08:43 · update #2
First: isolate x^2 by itself - add 729 to both sides (when you move a term to the opposite side, always use the opposite sign).
==> x^3 - 729+729 = 0+729
==> x^3 = 0+729
==> x^3 = 729
Sec: eliminate the exponent (find the cube root of both sides).
==> cube root√x^3 = +/- cube root√729
==> x = +/- cube root √729
Third: express 729 in simplest terms.
==> x = +/- cube root √9*9*9
==> x = +/- 9
==> x = -9, 9
*Check to make sure both work - replace each one with "x" in the equation. The only solution is 9.
2007-11-24 13:35:41 · answer #1 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
1. Add 729 to both sides to get:
x^3=729
2. Then, take the cube root of both sides:
x=9
2007-11-24 19:05:24 · answer #2 · answered by Jay Jay 2 · 0⤊ 0⤋
add -729 to both sides
x^3 = 729
Take the 3rd root of both sides. Use a calculator to take the 3rd rood of 729
x = 9
2007-11-24 19:05:47 · answer #3 · answered by Mohsin 3 · 0⤊ 0⤋
x^3 - 729 = 0
x^3 = 729
x = 9
.
2007-11-24 19:05:54 · answer #4 · answered by Robert L 7 · 0⤊ 0⤋
x^3-729 = (x-9)(x^2 +9x +81)
linear factor gives x = 9, the real root
quadratic gives ( -9 +- sqrt(81-324))/2 = (-9 +-9sqrt(-3))/2
= -4.5 +- 4.5sqrt(3)i, the conjugate pair of complex roots
2007-11-24 19:10:49 · answer #5 · answered by holdm 7 · 0⤊ 1⤋
it's easy
instead of doing your homework
here is a hint
1. move all the numbers to one side of the equal sign
2. make sure the ohther side of the equal sign is the x variable
3. solve for x
2007-11-24 19:04:53 · answer #6 · answered by Clueless 5 · 0⤊ 0⤋