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Show that the roots of the quadratic eqution: (b-c)x^2 + (c-a)x + (a-b)=0 are equal if c+a=2b

Posted by Arpit Arpit 5 years, 8 months ago

CBSE > Class 10 > Mathematics

3 answers

Nemha Chala 5 years, 8 months ago

b^2-4ac=0 (c-a)^2-4(b-c)(a-b)=0 c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0 a^2+4b^2+c^2-4ab-4bc+2ac=0 a^2+(-2b)^2+c^2+2×a×(-2b)+2×(-2b)+2×(-2b)×c+2×a×c=0 {a+(-2b)+c}^2=0 a-2ab+c=0 2b=a+c Hence,proved

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Anshul Soni 5 years, 8 months ago

If roots are equal Then we have to prove that B^2-4AC=0 =(c-a) ^2-4(b-c) (a-b) =c^2+a^2-2ac-4(ab-b^2-ac+bc) =a^2+c^2-2ac-4ab+4b^2+4ac-4bc =a^2+4b^2+c^2-4ab-4bc+2ac =(a-2b+c)^2. (eq1) But given that c+a=2b Then a-2b+c=0. (eq2) So, by (1) and (2) (0)^2 =0 So, B^2-4AC=0 So., the equation has two equal roots

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Sarmeet Virk 5 years, 8 months ago

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CBSE > Class 10 > Mathematics