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  4. Both the roots of the equation (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0 are always

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Q. Both the roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always

IIT JEEIIT JEE 1980Complex Numbers and Quadratic Equations

Solution:

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
$\Rightarrow 3x^2-2(a+b+c)x+(ab+bc+ca)=0$
Now, discriminant $=4(a+b+c)^2-12(ab+bc+ca)$
$=4 \{a^2+b^2+c^2-ab-bc-ca\}$
$=2 \{(a-b)^2+(b-c)^2+(c-a)^2\}$
which is always positive.
Hence, both roots are real.