Question

Let $f\left(x\right)=x^3+2x^2+3x+4$, then the equation $\frac{1}{x-f\left(1\right)}+\frac{2}{x-f\left(2\right)}+\frac{3}{x-f\left(3\right)}=0$, has

• A. 1 real root
• B. 2 real roots
• C. all three roots are real
• D. no real root exist

Answer 1

The correct option is B 2 real roots

$f\left(x\right)=x^3+2x^2+3x+4$
Then
$f\left(1\right)=1^3+{2.1}^2+3.1+4=1+2+3+4=10f\left(2\right)=2^3+{2.2}^2+3.2+4=8+8+6+4=26f\left(3\right)=3^3+{2.3}^2+3.3+4=27+18+9+4=56$

$\frac{1}{x-f\left(1\right)}+\frac{2}{x-f\left(2\right)}+\frac{3}{x-f\left(3\right)}=0$

$\Rightarrow 1(x-f\left(2\right))(x-f\left(3\right))+2(x-f\left(1\right))(x-f\left(3\right))+3(x-f\left(1\right))(x-f\left(2\right))=0$
$\Rightarrow (x-26)(x-56)+2(x-10)(x-56)+3(x-10)(x-26)=0$
$\Rightarrow 6x^2-322x+3356=0$

Comparing with General form of quadratic equation

Here, $a=6,b=-322,c=3356$

$\therefore D=b^2-4ac={322}^2-4\times 6\times 3356=23140>0$

Therefore, this quadratic equation have $2$ real roots