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Question

If the roots of the equation (x+a)(x+b)+(x+a)(x+c)+(x+b)(x+c)=0 are equal, then

A
(a+b+c)2ab+bc+ca
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B
(a+b+c)2<ab+bc+ca
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C
(a+b+c)2=ab+bc+ca
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D
None of these
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Solution

The correct option is C (a+b+c)2=ab+bc+ca
(x+a)(x+b)+(x+a)(x+c)+(x+b)(x+c)=0x2+(a+b)x+ab+x2+(a+c)x+ac+x2+(b+c)x+bc=03x2+2(a+b+c)x+ab+bc+ac)=0
roots are equal so discriminant is 0
D=0(2(a+b+c))24.3(ab+bc+ac)=04(a+b+c)24.3(ab+bc+ac)=0a2+b2+c2+2(ab+bc+ac)3(ab+bc+ac)=0a2+b2+c2=ab+bc+ac

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