((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 (and not 0 as it was written in previous variant, im sorry :()
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 21:54:47 · 5 answers · asked by lika n 1 in Science & Mathematics ➔ Mathematics
it is the same as writing
x^2 - x^2 + 1 = 1
2006-08-22 22:09:21 · answer #1 · answered by camedamdan 2 · 0⤊ 1⤋
Because this is not equation but identity and works for any x
2006-08-23 07:08:59 · answer #2 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
there is a zero divided by zero!
2006-08-23 05:13:54 · answer #3 · answered by blind_chameleon 5 · 0⤊ 0⤋
Confusing...I give up
2006-08-23 05:18:12 · answer #4 · answered by Julian 3 · 0⤊ 1⤋
(m-n) = -(n-m)
rewrite the equation:
(x-a)(x-b)/-(a-c)*-(b-c) = (x-a)(x-b)/(a-c)(b-c)
(x-b)(x-c)/(a-b)(a-c)
(x-c)(x-a)/(b-c)*-(a-b) = - (x-c)(x-a)/(b-c)(a-b)
LCM = (a-b)(a-c)(b-c)
((x-a)(x-b)(a-b) + (x-b)(x-c)(b-c) - (x-c)(x-a)(a-c))/(a-b)(a-c)(b-c) = 1
(x-a)(x-b)(a-b) + (x-b)(x-c)(b-c) - (x-c)(x-a)(a-c) = (a-b)(a-c)(b-c)
(x-a)(x-b)(a-b) + (x-b)(x-c)(b-c) - (x-c)(x-a)(a-c) - (a-b)(a-c)(b-c) =0
(xx-ax-bx+ab)(a-b) + (xx-bx-cx+bc)(b-c) - (xx-cx-ax+ac)(a-c) - (aa-ab-ac+bc)(b-c) = 0
axx-aax-abx+aab-bxx+abx+bbx-abb+bxx-bbx-bcx+bbc-cxx+bcx+ccx-bcc-axx+acx+aax-aac+cxx-ccx-acx+acc-aab+abb+abc-bbc+aac-abc-acc+bcc=0
Grouping the items:
axx-bxx+bxx-cxx-axx+cxx = 0
-aax-abx+abx+bbx-bbx-bcx+bcx+ccx+acx+aax-ccx-acx = 0
+aab-abb+bbc-bcc-aac+acc-aab+abb+abc-bbc+aac-abc-acc+bcc = 0
You have 0x^2 + 0x + 0 = 0
2006-08-23 05:35:48 · answer #5 · answered by Anonymous · 0⤊ 0⤋