Both the roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always
- positive
- negative
- real
- None of these
The Correct Option is C
Solution and Explanation
$\Rightarrow 3x^2-2(a+b+c)x+(ab+bc+ca)=0$
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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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
