Question

For the polynomial equation
$\frac{(x+q)(x+r)}{(q-p)(r-p)}+\frac{(x+r)(x+p)}{(r-q)(p-q)}+\frac{(x+p)(x+q)}{(p-r)(q-r)}-1=0$, where $p,q,r$ are distinct real numbers.
Which of the following statement is correct?

• A. It has $2$ distinct real roots.
• B. It has no real roots.
• C. It has real and equal roots.
• D. It has more than $2$ real roots.

Answer 1

The correct option is D It has more than $2$ real roots.

$\frac{(x+q)(x+r)}{(q-p)(r-p)}+\frac{(x+r)(x+p)}{(r-q)(p-q)}+\frac{(x+p)(x+q)}{(p-r)(q-r)}-1=0$
Putting $x=-q$
$\frac{(-q+r)(-q+p)}{(r-q)(p-q)}-1=0\Rightarrow 1-1=0$
Similarly,
$x=-r,-p$ are also the roots of the polynomial.
Therefore, the given equation is an identity, so it has more than $2$ roots.