((x-a)(x-b))/((c-a)(c-b))+
+((x-b)(x-c))/((a-b)(a-c))+
+((x-c)(x-a))/((b-c)(b-a))=
=0
u can see that its a 2nd degree equation for x, but how come that it has 3 roots a,b,c? i mean if u put in the place of x any of them equetion will be true
2006-08-22 21:32:33 · 5 answers · asked by lika n 1 in Science & Mathematics ➔ Mathematics
it is not true
i.e.
put a in place of x
((x-a)(x-b))/((c-a)(c-b)) ------------>0
((x-b)(x-c))/((a-b)(a-c)) ------------>1
((x-c)(x-a))/((b-c)(b-a)) ------------>0
0+1+0 = 1
see?
none of a,b,c are solutions for this equation
2006-08-22 21:38:12 · answer #1 · answered by camedamdan 2 · 0⤊ 0⤋
With all this algebraic "stuff"....did you forget to add your fractions like a good 6th grade student and list the values for which the denominator is 0 ?????
After you do the arithmetic...before you get to the algebra...you get a quadratic equation....2 answers
2006-08-22 21:43:47 · answer #2 · answered by Gemelli2 5 · 0⤊ 0⤋
Nice try,dear daughter I think you didn't write the paradox correctly
It's ((x-a)(x-b))/((c-a)(c-b))+((x-b)(x-c))/((a-b)(a-c))
+((x-c)(x-a))/((b-c)(b-a))=1
the solution is so simple..if you tried to simplify this formula to a polynomial you will get 1=1!!
so this is NOT a second degree equation It's not an equation at
all..you can do the same game with any degree and any number
of arbitrary constants a,b,c,d................ etc
2006-08-22 22:05:09 · answer #3 · answered by mohamed.kapci 3 · 0⤊ 1⤋
The last pair 'a docs I saw couldn't agree with each other. What a conundrum!
2006-08-22 21:44:55 · answer #4 · answered by MaqAtak 4 · 0⤊ 0⤋
Can't get you.
But I dont like paradoxes.
2006-08-22 21:35:01 · answer #5 · answered by Ω Nookey™ 7 · 0⤊ 0⤋