3p^4= 5p^2 - 2
Solve
2007-08-23 07:17:01 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3p^4-5p^2 + 2 = 0
(3p^2-2)(p^2-1) = 0
p = ±√(2/3), ±1
2007-08-23 07:20:46 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
3p^4=5p^2-2
or 3p^4-5p^2+2=0
or 3p^4-3p^2-2p^2+2=0
or 3p^2(p^2-1)-2(p^2-1)=0
or (3p^2-2)(p^2-1)=0
so either 3p^2-2=0 or p^2-1=0
so p^2=2/3 or p^2=1
so p=+-sq rt(2/3) or p=+-1 ans
2007-08-23 14:31:14 · answer #2 · answered by MAHAANIM07 4 · 0⤊ 0⤋
rewrite the eqn:
3p^4-5p^2+2=0
use the quadratic eqn to solve:
[-b +/- sqrt(b^2-4ac)]/2a
where a = 3, b = -5, c = 2.
You will have to distinct answers..
2007-08-23 14:21:03 · answer #3 · answered by miggitymaggz 5 · 0⤊ 1⤋
let x = p². then
3x² - 5x + 2 = 0
(3x - 2)(x - 1) = 0
x = 2/3 or x = 1
p² = 2/3 or p² = 1
p = ±(â6)/3 or p = ±1.
2007-08-23 14:23:07 · answer #4 · answered by Philo 7 · 0⤊ 0⤋