2006-12-02 06:53:00 · 8 answers · asked by luvintheharley 1 in Science & Mathematics ➔ Mathematics
break up (x-8)^2 first:
(x-8)(x-8) = x^2 -16x + 64 = -64
x^2 - 16x + 128 = 0
then use the quadratic equation:
x = [-b +/- sqrt(b^2 - 4ac)]/4ac
and x = 8 + 8i and 8 - 8i
2006-12-02 08:05:57 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Is this an imaginary number problem? If not, then there's no solution. If it is then take the square root of both sides. Then you get 8i on the right and x-8 on the left. Then just add the 8 to the right side to get x=8i +8
2006-12-02 15:00:26 · answer #2 · answered by hmmm 2 · 2⤊ 0⤋
( x - 8 )^2 = -64
taking sqrt on both side
you wil get,
x - 8 = -8i or x - 8 = +8i
// since the sqtr of -ve no is an imaginary part. the sign may be either +ve or -ve
x = 8 - 8i or x = 8 + 8i
thats the solution.
2006-12-02 15:13:22 · answer #3 · answered by candy 2 · 1⤊ 0⤋
Whoa! Sorry, I tried, but after ripping up several sheets of scrap paper, I have determined nothing except that I forgot how to do these buggers.
2006-12-02 15:06:10 · answer #4 · answered by chameleon 3 · 0⤊ 0⤋
you are only half right
( X - 8 )^@
= X^2 + 64
y
you muli. all the variables and numbers insidr the parenterer by the,selfs and when you mult. like sign as in two neg. signs you get a pos. answer !!!!
2006-12-02 14:59:01 · answer #5 · answered by Anonymous · 0⤊ 2⤋
take the square root of both sides
then solve for x
2006-12-02 14:57:08 · answer #6 · answered by The Old Professor 5 · 0⤊ 1⤋
No Solution
2006-12-02 15:04:41 · answer #7 · answered by Anonymous · 0⤊ 2⤋
are you doing complex numbers, or was that a mistake?
2006-12-02 15:00:03 · answer #8 · answered by averagebear 6 · 1⤊ 1⤋