2007-12-16 08:20:23 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x ² + 2 x + 6 = 0
x = [- 2 ± √(4 - 24) ] / 2
x = [ - 2 ± √(20 i ²) ] / 2
x = [ - 2 ± 2 i √5 ] / 2
x = - 1 ± i √5
2007-12-20 05:27:19 · answer #1 · answered by Como 7 · 1⤊ 0⤋
Using the Quadratic Formula to solve
1x²+2x+6
a=1
b=2
c=6
x = [-b ± â(b²-4ac)]/(2a)
= [-(2) ± â(2² - 4(1)(6))]/2(1)
= [-2 ± â(-20)]/2
= [-2 ± â4â(-1)â5]/2
= [-2 ± 2iâ5]/2
= -1 ± iâ5
x₊ = -1+ 2.236i
x₋ = -1- 2.236i
2007-12-16 08:25:49 · answer #2 · answered by DWRead 7 · 0⤊ 0⤋
y = x^2 + 2x + 6. If we rewrite it as y = ax^2 + bx + c, we have a = 1, b = 2, and c = 6. The quadratic formula is x = (-b +/- sqrt(b^2 - 4ac)) / 2a, which provides the values of x such that y = 0. In this case, x = (-2 +/- sqrt(2^2 - 4*1*6)) / (2*1) = (-2 +/- sqrt(4 - 24)) / 2 = (-2 +/- sqrt(-20)) / 2 = (-2 +/- 2i*sqrt(5)) / 2 = -1 +/- i*sqrt(5), which is a complex conjugate pair.
2007-12-16 08:23:15 · answer #3 · answered by DavidK93 7 · 0⤊ 1⤋
a=1 b=2 c=6
-b + or - (square root of) b squared - 4ac divided by 2a
answer is -1+or - i root 5
2007-12-16 08:28:50 · answer #4 · answered by Music=Life 2 · 0⤊ 0⤋
Let's write it with ^2 for squared:
x^2 + 2x + 6
a = 1
b = 2
c = 6
x = [-2 +/- sqrt(2^2 - 4(1)(6))]/2(1)
= [-2 +/- sqrt(4-24)]/2
= [-2 +/- sqrt(-20)]/2
= [-2 +/- 2sqrt(-5)]/2
= -1 +/- i sqrt(5)
That's it! :)
2007-12-16 08:26:22 · answer #5 · answered by Marley K 7 · 0⤊ 0⤋
x1 = -1+root(4-24) = -1+i*root(20)
x2 = -1-root(4-24) = -1 -i*root(20)
2007-12-16 08:24:54 · answer #6 · answered by Nur S 4 · 0⤊ 0⤋
Do your own homework. (My way of saying...I am a retard)
2007-12-16 08:27:58 · answer #7 · answered by Why ask me? 4 · 0⤊ 0⤋