Question

For all real values of $p$ , the roots of the equation
$\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-p}=0$ are

• A. real and distinct
• B. real and equal
• C. real
• D. imaginary

Answer 1

The correct option is A real and distinct

$\frac{(x-1)(x-p)+x(x-p)+x(x-1)}{x(x-1)(x-p)}=0$
$\Rightarrow 3x^2-(1+p+p+1)x+p=0$
$3x^2-2(1+p)x+p=0$
$D=b^2-4ac=\left(2\right(1+p){)}^2-4(3p)$
$=4(1+p^2+2p-3p)$$=4(1+p^2-p)$
$=4(p^2-p+\frac{1}{4}+\frac{3}{4})$
$=4({(p-\frac{1}{2})}^2+\frac{3}{4})$
$\therefore D>0$
$\therefore$ Roots are real and distinct .