Question

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

• A. $(a+b+c{)}^2\ge ab+bc+ca$
• B. $(a+b+c{)}^2<ab+bc+ca$
• C. $(a+b+c{)}^2=ab+bc+ca$
• D. None of these

Answer 1

The correct option is C $(a+b+c{)}^2=ab+bc+ca$

$(x+a)(x+b)+(x+a)(x+c)+(x+b)(x+c)=0x^2+(a+b)x+ab+x^2+(a+c)x+ac+x^2+(b+c)x+bc=03x^2+2(a+b+c)x+ab+bc+ac)=0$

roots are equal so discriminant is 0

$D=0{\left(2\right(a+b+c\left)\right)}^2-4.3(ab+bc+ac)=04{(a+b+c)}^2-4.3(ab+bc+ac)=0a^2+b^2+c^2+2(ab+bc+ac)-3(ab+bc+ac)=0a^2+b^2+c^2=ab+bc+ac$