gimme the value of x
2006-10-30 17:00:12 · 15 answers · asked by KiraYammi 2 in Science & Mathematics ➔ Mathematics
x^2-123456789+10=0
x^2-123456779=0
x^2=123456779
x=11111.1106 or -11111.1106
2006-10-30 17:15:39 · answer #1 · answered by AD 2 · 0⤊ 0⤋
X^2-123456789+10=0
x^2=123456779
x=+/-111111.1106...
2006-10-31 09:48:17 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋
if the x2 is meant to be x^2, then there are 2 recommendations. component x^2 + 7x + 10 = 0 to get (x + 5) (x + 2) = 0 set each and each component = 0 --> (x + 5) = 0 --> x = -5 and (x + 2) = 0 --? x = -2 So x = -5 and x = -2 are the two recommendations
2016-12-16 16:59:11 · answer #3 · answered by vanderlinden 3 · 0⤊ 0⤋
x^2-123456789+10=0
x^2=123456789-10
x^2=123456779
x= sqrt (123456779)
x = 11111.110610555545 , x = (- 11111.110610555545 )
2006-10-30 17:19:43 · answer #4 · answered by safrodin 3 · 0⤊ 0⤋
x^2=123456789-10=123456779
so x=square root of 123456779
or x=-square root of 123456779
2006-10-30 18:04:33 · answer #5 · answered by peterwan1982 2 · 0⤊ 0⤋
x^2=123456789-10
x^2=123456779
x^2=11111.11061
2006-10-30 21:46:53 · answer #6 · answered by amila 1 · 0⤊ 0⤋
Is the linear term a coefficient or not ? If not, then the solution is the + or - the square root of 123456779.
2006-10-30 17:10:45 · answer #7 · answered by mathlete1 3 · 0⤊ 0⤋
Pizza.
:)
2006-10-30 21:06:05 · answer #8 · answered by CJ 3 · 0⤊ 0⤋
write the question properly
2006-10-30 17:30:00 · answer #9 · answered by Nick 3 · 0⤊ 0⤋
+ or - the square root of (123456779)
or about + or - 11111.11061
2006-10-30 17:03:21 · answer #10 · answered by ryan w 2 · 0⤊ 0⤋