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


A) positive

B) negative

C) real

D) imaginary

Correct Answer:
C) real

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.

Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations








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