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

positive

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real

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negative

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none of these

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Solution

The correct option is B real
$(x - b) (x - c) + (x - a) (x - c) + (x - a) (x - b) = 0$

$\Rightarrow x^{2} - 2 x (a + b + c) + a b + b c + a c = 0$

Now, $D = 4 [{(a + b + c)}^{2} - a b - b c - a c] = 4 [a^{2} + b^{2} + c^{2} + a b + b c + a c]$

$D = 2 [2 a^{2} + 2 b^{2} + 2 c^{2} + 2 a b + 2 b c + 2 c a] \Rightarrow D = 2 [a^{2} + b^{2} + 2 a b + b^{2} + c^{2} + 2 b c + c^{2} + a^{2} + 2 c a]$

$\Rightarrow D = 2 [{(a + b)}^{2} + {(b + c)}^{2} + {(a + c)}^{2}] > 0$

Therefore, roots are real.
Ans: B

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The correct option is B real(x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=0⇒x2−2x(a+b+c)+ab+bc+ac=0Now, D=4[(a+b+c)2−ab−bc−ac]=4[a2+b2+c2+ab+bc+ac]D=2[2a2+2b2+2c2+2ab+2bc+2ca]⇒D=2[a2+b2+2ab+b2+c2+2bc+c2+a2+2ca]⇒D=2[(a+b)2+(b+c)2+(a+c)2]>0Therefore, roots are real.Ans: B

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