if the roots of the quadratic equation
(a-b)x^2 +(b-c)x +(c-a)=0 are equal ,then prove that 2a=b+c
shadowsabers03:
Proved a lot!!!
Answers
Answered by
8
Roots are equal. So discriminant is zero.
Therefore,
[tex](b-c)^2-(4(a-b)(c-a)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ac-a^2-bc+ab)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ab-bc+ac-a^2)) = 0 \\ \\ (b^2-2bc+c^2)-(4ab-4bc+4ac-4a^2)=0 \\ \\ b^2-2bc+c^2-4ab+4bc-4ac+4a^2=0 \\ \\ 4a^2+b^2+c^2-4ab+2bc-4ac=0 \\ \\ (2a-b-c)^2=0 \\ \\ 2a-b-c=0 \\ \\ 2a=b+c [/tex]
Hence proved!
Thank you. Have a nice day. :-)
#adithyasajeevan
Third time I'm proving this!!! ;-)
sorry I cant understand the 6th step
I rearranged the terms there.
I brought 4a^2 first, then b^2, c^2 and others in that order. -2bc and 4bc are added together, to get 2bc.
Answered by
1
Heya!!!
Answer to your question:
For equal roots,
b^2-4ac=0
(b-c)^2-4 (a-b)(c-a)=0
b^2 +c^2 +2bc -4 (ac - a^2 -bc +ab)=0
b^2 +c^2 +2bc -4ac +4a^2 +4bc -4ab=0
(2a)^2+(-b)^2-(c)^2 +2 (2a)(-b) +2 (-b)(-c) +2(2a)(-c)=0
(2a-b-c)^2=0
2a-b-c=0
2a=b+c
Hopeit helps ^_^
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