z^8 - 14z^4 + 16 = 0
2006-11-26 14:11:19 · 7 answers · asked by alchem 1 in Science & Mathematics ➔ Mathematics
(z^4)^2 - 14z^4 +16 =0
Quadratic factoring:
z^4 = (14+sqrt(196-64))/2 = 7+sqrt(33) =
12.7446 so z = 1.89 or -1.89
OR
z^4 = 7-sqrt(33)= 1.2554
so z= 1.059 or -1.059
2006-11-26 14:30:23 · answer #1 · answered by Jimbo 5 · 0⤊ 0⤋
are you sure you wrote the problem correctly? The solution would involve complicated decimals. A similar equation that would factor properly is z^8 - 17z^4 + 16 = 0. The solutions would be z = 2 and z = 1[Problem factors into (z^4 - 16)(z^4 - 1)]
2006-11-26 22:21:38 · answer #2 · answered by Tesline T 2 · 0⤊ 0⤋
z^8 - 14z^4 + 16 = 0
put z^4=x
x^2-14x+16=0
x^2-14x+16=0
solve for x
and then take
z^4=x
z=x^1/4
2006-11-26 22:14:34 · answer #3 · answered by Dupinder jeet kaur k 2 · 0⤊ 0⤋
You have to introduce a new variable: Z= z^4
So your equation becomes: Z^2 -14 Z +16 =0
Solve this, you'll find two solutions for Z. Then for each solution use the definition of the variable (Z=z^4) and you'll find z.
Good luck!
2006-11-26 22:16:57 · answer #4 · answered by Dragonus 2 · 0⤊ 0⤋
you can use factoring to solve this problem. If you have a Texas Instrument calculator (Ti 83, 84, or 89 would work the best) you can see if you have POLYSMLT. It is a brilliant program that will solve this for you. The solution to this is +/- 1.058519253 and
+/- 1.889431859
2006-11-26 22:17:35 · answer #5 · answered by elderly woman behind the counter 2 · 0⤊ 0⤋
It becomes a quadratic equation when you replace z^4 with x.
Solve the quadratic equation.
You will get two values for x say x1 and x2
Then solve z^4 = x2 and z^4 = x2
2006-11-26 22:18:43 · answer #6 · answered by sudhir49garg 2 · 0⤊ 0⤋
try the substitution x=z^4
that should get you on your way
2006-11-26 22:14:46 · answer #7 · answered by Anonymous · 0⤊ 0⤋