b^4+13b^2+36=0
2007-10-22 04:33:10 · 7 answers · asked by mormar 1 in Science & Mathematics ➔ Mathematics
b^2
x^2 +13X +36 = 0
(x+9)(x+4) = 0
x = -9 and x = -4
b^2 = -9 --> b = +/-3i
b^2 = -4 --> b = +/- 2i
2007-10-22 04:57:42 · answer #1 · answered by sumguy 1 · 0⤊ 0⤋
b^4+13b^2 + 36=0
b^2=y
y^2+13y +36=0
(y+9) (y+4) = 0
Y=-9 or Y =-4
Then b^2 = y
b^2 = -9
b = +or -3i
b ^=-4
b =+or -2 i
2007-10-22 04:47:28 · answer #2 · answered by xmashi 3 · 0⤊ 0⤋
b^4+13b^2+36=0
(b^2+9)(b^2+4) = 0
b^2 = -9 or -4
When b^2 = -9, b = +/- 3i
When b^2 = -4, b = +/- 2i
2007-10-22 04:39:05 · answer #3 · answered by cllau74 4 · 0⤊ 0⤋
Let x = b^2
Then x^2 +13X +36 = 0
(x+9)(x+4) = 0
x = -9 and x = -4
Thus b^2 = -9 --> b = +/-3i
and b^2 = -4 --> b = +/- 2i
2007-10-22 04:39:32 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
factor it: (b^2 + 9)(b^2 + 4) = 0
set each factor to =
b^2 + 9 = 0 or b^2 + 4 = 0
b^2 = -9 or b^2 = -4
take the square root
b = +/- 3i or b = +/- 2i
2007-10-22 04:38:58 · answer #5 · answered by Linda K 5 · 0⤊ 0⤋
(b ² + 9)(b² + 4) = 0
b² + 9 = 0 , b² + 4 = 0
b² = - 9 , b² = - 4
b = ± 3 i , b = ± 2 i
2007-10-22 07:01:10 · answer #6 · answered by Como 7 · 0⤊ 0⤋
i think that question belongs in "marraige and divorce."
2007-10-22 04:40:44 · answer #7 · answered by Anonymous · 0⤊ 0⤋