Both the roots of the given equation. \( \Large \left(x-a\right) \left(x-b\right)+ \left(x-b\right) \left(x-c\right)+ \left(x-c\right) \left(x-a\right)=0 \) are always
Correct Answer: Description for Correct answer:
Given equation:
\( \Large \left(x-a\right) \left(x-b\right)+ \left(x-b\right) \left(x-c\right)+ \left(x-c\right) \left(x-a\right)=0 \)
\( \Large 3x^{2}-2 \left(a+b+c\right)x+ \left(ab+bc+ca\right)=0 \)
Now, \( \Large BZ-4AC=4\left[ \left(a+b+c+\right)^{2}-3 \left(ab+bc+ca\right) \right] \)
= \( \Large 4 \left(a^{2}+b^{2}+c^{2}-ab-bc-ac\right) \)
= \( \Large 2\{ \left(a-b\right)^{2}+ \left(b-c\right)^{2}+ \left(c-a\right)^{2} \}\ge 0 \)
Hence, both roots are always real.
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