Prove that if
ac-b2 bd-c2
------ = ------
a-2b-c b-2c+d
then both of the fractions are also equal to
ad-bc
-------
a-b-c+d
(Assuming no divisions by zero, and b≠c.)
There is some error in the formulation of this. I offer an example that shows the error.
Suppose a=2 b=6 c=4 d=3
Then
(ac-b^2)/(a-2b-c) = (8-36)/(2-12-4)=-28/-14=2
(bd-c^2)/(b-2c+d) = (18-16)/(6-8+3)= 2/1=2
So, both fractions are equal, but:
(ad-bc)/(a-b-c+d) = (6-24)/(2-6-4+3)= -18/-5= 18/5
Differt from the two equal fractions.
This works only (with the example) if the numerator is ac-bd instead of ad-bc, or if the denominator of the third fraction is -a instead of a.
Edited on February 2, 2018, 5:45 am
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Posted by armando
on 2018-02-01 17:06:58 |
Surely not