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Both the roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always
Both the roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always
Updated On: Jun 14, 2022
- positive
- negative
- real
- None of these
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The Correct Option is C
Solution and Explanation
$(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$
$\Rightarrow 3x^2-2(a+b+c)x+(ab+bc+ca)=0$
Now, discriminant $=4(a+b+c)^2-12(ab+bc+ca)$
$=4\big(a^2+b^2+c^2-ab-bc-ca\big)$
$=2\Big\{(a-b)^2+(b-c)^2+(c-a)^2\Big\}$
which is always positive.
Hence, both roots are real.
$\Rightarrow 3x^2-2(a+b+c)x+(ab+bc+ca)=0$
Now, discriminant $=4(a+b+c)^2-12(ab+bc+ca)$
$=4\big(a^2+b^2+c^2-ab-bc-ca\big)$
$=2\Big\{(a-b)^2+(b-c)^2+(c-a)^2\Big\}$
which is always positive.
Hence, both roots are real.
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