Question

If $a,b,c$ are the roots of $x^3+4x+1=0$ then
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=$

• A. 2
• B. 3
• C. 4
• D. -4

Answer 1

The correct option is C 4

$a,b,c$ are the roots of $x^3+4x+1=0$

$a+b+c=-\frac{coeff.of{x}^2}{coeff.of{x}^3}=-\frac{0}{1}=0$

$ab+bc+ca=\frac{coeff.ofx}{coeff.of{x}^3}=\frac{4}{1}=4$

$abc=-\frac{constantterm}{coeff.x^3}=-\frac{1}{1}=-1$

Now,

$a+b+c=0$

$a+b=-c$

$b+c=-a$

$c+a=-b$

Now,

$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{-c}+\frac{1}{(-a)}+\frac{1}{(-b)}$

$=-(\frac{1}{c}+\frac{1}{a}+\frac{1}{b})$

$=-\left(\frac{ab+bc+ca}{abc}\right)$

$=-\left(\frac{4}{-1}\right)$

$=4$