2007-04-28 03:22:57 · 15 answers · asked by HV 1 in Science & Mathematics ➔ Mathematics
x^4-27x=0
x(x^3-27)=0
x=0 or x^3=3^3
x=0 or x=3
2007-04-28 03:28:15 · answer #1 · answered by Smarty 1 · 1⤊ 1⤋
Do you advise: x^4 - 5x^2 + 6 = 0 if so, the difficulty could desire to be solved this form x^4 - 5x^2 + 6 = 0 assume x^2 = y so : y^2 - 5y + 6 = 0 ( y - 2 )( y - 3 ) = 0 y1=3 y2=2 y^2 = x so x= root(y) x1 = root 3 x2 = - root 3 x3 = root 2 x4 = - root 2
2016-10-14 00:07:40 · answer #2 · answered by ? 4 · 0⤊ 0⤋
x^4 -27 x = 0
x^4 = 27 x
x^3 = 27
x = 3
2007-04-28 03:57:24 · answer #3 · answered by calvin o 5 · 0⤊ 0⤋
x^4-27x=0
=> x(x^3-27)=0
=>x{(x)^3-(3)^3}=0
=>x(x-3)(x^2+3x+9)=0
kTherefore either x=0 or x-3=0
Hence x=0 or 3
2007-04-28 03:33:00 · answer #4 · answered by alpha 7 · 0⤊ 0⤋
3
2007-04-28 03:26:15 · answer #5 · answered by Anonymous · 1⤊ 4⤋
put x in factors
x ( x^3-27 ) = 0
27 is the third power of three
so (x^3-27) = (x-3) ( x^2+3x +3)
and your expression is x^4-27 = x (x-3) (x^2+3x +3)
the roots are x=0 , x=+3 and it is all since the third factor is always positive
2007-04-28 03:36:04 · answer #6 · answered by maussy 7 · 0⤊ 0⤋
That would be 3
2007-04-28 03:28:02 · answer #7 · answered by Qrias 3 · 0⤊ 3⤋
x^4 -27x=0 factor out x
x(x^3-27)=0 difference of two cubes
x[(x-3)^3]=0
x=0 or 3
2007-04-28 03:28:15 · answer #8 · answered by Anonymous · 2⤊ 2⤋
x(x^3 - 27) = 0
x^3 - 27 = 0 x = 0 is one solution
x^3 = 27 take cube root
x = 3 is other solution
2007-04-28 03:28:45 · answer #9 · answered by Ben 1 · 1⤊ 0⤋
easy x= 0 and 3
2007-04-28 03:28:02 · answer #10 · answered by * SeñørITA * 6 · 3⤊ 0⤋