Math, asked by saibiradae06, 1 year ago

If roots of quadratic equation (a-b)x square+(b-c)x+(c-a)=0 are equal .Then prove that 2a=b+c


shadowsabers03: Please see my answer: https://brainly.in/question/8369571

Answers

Answered by Anonymous
12
I hope this will help you......

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saibiradae06: thanks buddy
Anonymous: no mention......
Answered by shadowsabers03
0

Roots are equal. So discriminant is zero.  

Therefore,  

(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

Hence proved!  

Thank you. Have a nice day. :-)  

#adithyasajeevan

           


shadowsabers03: Oh, I'm so exhausted because I'm proving the same equation fourth time!!!
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