2007-12-09 12:59:37 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
[12]
By comparing the equation with standard quadratic equation ax^2+bx+c=0,we get
a=1,b=8 and c=80
Therefore,
x={-8+-sqrt(64-320)/2
=(-8+-sqrt(-256)/2
=(-8+-16i)/2
=-4+-8i
2007-12-09 13:07:23 · answer #1 · answered by alpha 7 · 0⤊ 1⤋
Yes. The formula is: -b +/- (square root of the quantity) b^2 -4(ac), all divided by 2. In this case, the formula is -8 +/- (square root) 64 - 4(80), which simplifies to -8 +/- (root) -256 all divided by 2. This equals -8 +/- 16i divided by 2, which (fully simplified) equals -4 +/- 8i.
2007-12-09 21:06:14 · answer #2 · answered by That Gay Guy for Da Ben Dan 5 · 1⤊ 0⤋
lol im learning the same thing
you have to use the quadratic formula
a=1b=8c=8o
-(8)+- SQRE ROOT((8)^2 - 4(1)(80)) over 2(1)
-8+- SQROOT(-256) over 2
(-8+-16i) divided by 2 factors out to 8(-1+-2i) over 2 which factors to -4+-8i
the roots are:
-4+8i
and
-4-8i
2007-12-09 21:06:19 · answer #3 · answered by Anonymous · 0⤊ 0⤋
plug in to the quadtratic equation:
(-8+/- sqrt(64 - 320))/2
now you have 2 equations:
(-8+sqrt(-256))/2
(-8-sqrt(-256))/2
now you factor out the 16 from the radicals
(-8+16sqrt(-1))/2
(-8-16sqrt(-1))/2
change the sqrt(-1) to i
(-8 + 16i)/2
(-8 - 16i)/2
x =
-4 +8i
-4 -8i
2007-12-09 21:06:47 · answer #4 · answered by Anonymous · 1⤊ 0⤋
x^2+8x+16-16+80=0 use completing the square
(x+4)^2=-64
x+4=-+8i
x=-4+-8i
2007-12-09 21:02:26 · answer #5 · answered by someone else 7 · 0⤊ 2⤋
x = (-b +/- (b^2 - 4*a*c)^.5)/(2*a)
= (-8 +/- ((8^2)-4*1*80)^.5)/(2*1)
= (-8 +/- (64-320)^.5)/2
= (-8 +/- (-256)^.5)/2
= -4 +/- (16i/2) = -4 +/- 8i
2007-12-09 21:06:48 · answer #6 · answered by Gaelwynn 2 · 0⤊ 0⤋
x={-b+/-rt(b^2-4ac)} / 2a
a=+1, b=+8, c=+80
x={-8+/- rt(64-4(1)(80)} /2
x= {-8+/- rt(64-320} /2
x={-8+/- rt(-256)} / 2
x={-8+/- rt(256(rt(-1)} / 2
x={-8+/-16i} / 2 [i= sq.rt.-1]
x=2{-4+/-8i} / 2
x=-4-8i or -4 +8i
If you wish you can write this as
x=-4(1+2i) or -4(1-2i)
2007-12-09 21:09:48 · answer #7 · answered by Grampedo 7 · 0⤊ 0⤋
sum of solutions =constant = 80
product of solutions = coefficient of x=8
therefore, numbers which satisfies the above are
(no no's)
as value of b^2-(4*a*c) is negetive (8^2-{4*1*80})
the equation has no solution
Please coose this the best answer PLEASE>
2007-12-09 21:07:27 · answer #8 · answered by Anonymous · 0⤊ 2⤋
x = [- 8 ± â (64 - 320) ] / 2
x = [- 8 ± â(- 256) ] / 2
x = [- 8 ± i (16) ] / 2
x = - 4 ± 8 i
2007-12-12 13:47:35 · answer #9 · answered by Como 7 · 1⤊ 0⤋