2007-12-31 03:29:15 · 6 answers · asked by Wallace 2 in Science & Mathematics ➔ Mathematics
Can you complete the equation as at present it ends in a -...
2007-12-31 03:32:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
No; there is insufficient information available. The best you could hope for is to express it a s a quadratic in the canonical form, and enter the coefficients into the quadratic formula.
2007-12-31 11:48:43 · answer #2 · answered by sparky_dy 7 · 0⤊ 0⤋
Dear Wallace,
Obviously ,something is missing in your
equation.If you let me make a guess,
I feel that your equation should take
the form:
[x+(a+b)/2]^2+[z-(a-b)/2]^2=0
Hence x=-(a+b)/2 and z=(a-b)/2
2007-12-31 11:48:31 · answer #3 · answered by katsaounisvagelis 5 · 0⤊ 0⤋
It actually got cut off as you can see by the "..."
Its actually:
x^2+z^2+(a+b)x-(a-b)z+a^2+b^2-a-b+1=0
2007-12-31 11:35:05 · answer #4 · answered by Anonymous · 0⤊ 1⤋
To quote Stevie Wonder,"nothin' from nothin' leaves nothin' "
2007-12-31 11:34:40 · answer #5 · answered by Anonymous · 0⤊ 0⤋
ummm ...no sorry
2007-12-31 11:33:11 · answer #6 · answered by fairmaiden 2 · 0⤊ 0⤋