Question

Consider the equation $\frac{1}{x+p}+\frac{1}{x+q}+\frac{1}{x+r}=\frac{3}{x}$

(Where p, q, r are distinct real numbers)

3. The number of real roots of the given equation is(are)

4. If p, q, $r \in R^+$ the number of positive real roots of given equation is(are)

• A. Two real roots (if $p+q+r\ne0$); One real root (if $p+q+r=0$)
• B. 0

Answer 1

Answer: (A), (B)

3. Two real roots (if $p+q+r\ne0$); One real root (if $p+q+r=0$).

4. For $p,q,r>0$: Number of positive real roots is 0.

Answer 2

We start with

$$\frac{1}{x+p}+\frac{1}{x+q}+\frac{1}{x+r}=\frac{3}{x},$$

with $p,q,r$ distinct real numbers and with the understanding that $x\ne 0$ and $x\ne -p,-q,-r$.

Step 1. Clear Denominators

Multiply both sides by

$$x(x+p)(x+q)(x+r)$$

to obtain

$$x\,(x+q)(x+r) + x\,(x+p)(x+r) + x\,(x+p)(x+q) = 3\,(x+p)(x+q)(x+r).$$

Step 2. Expand the Left‐Hand Side (LHS)

Notice that:

$$(x+q)(x+r)= x^2+(q+r)x+qr,$$ $$(x+p)(x+r)= x^2+(p+r)x+pr,$$ $$(x+p)(x+q)= x^2+(p+q)x+pq.$$

Thus,

$$\text{LHS} = x\Bigl[\,3x^2 + \bigl((p+q)+(p+r)+(q+r)\bigr)x + (pq+pr+qr) \Bigr].$$

Since

$$(p+q)+(p+r)+(q+r)= 2(p+q+r),$$

we have

$$\text{LHS} = 3x^3 + 2(p+q+r)x^2 + (pq+pr+qr)x.$$

Step 3. Expand the Right‐Hand Side (RHS)

Expand

$$(x+p)(x+q)(x+r) = x^3+(p+q+r)x^2+(pq+pr+qr)x+pqr,$$

so that

$$\text{RHS} = 3x^3 + 3(p+q+r)x^2+3(pq+pr+qr)x+3pqr.$$

Step 4. Form the Reduced Equation

Subtract the RHS from the LHS:

$$[3x^3+2(p+q+r)x^2+(pq+pr+qr)x] - [3x^3+3(p+q+r)x^2+3(pq+pr+qr)x+3pqr] = 0.$$

Cancelling $3x^3$ we get:

$$- (p+q+r)x^2 -2(pq+pr+qr)x - 3pqr= 0.$$

Multiply through by $-1$:

$$(p+q+r)x^2+2(pq+pr+qr)x+3pqr=0.$$

This is a quadratic in $x$.

Answer to (3):
In general, if $p+q+r\ne 0$ (the generic situation), this quadratic has two real roots (its discriminant

$$D=4\Bigl[(pq+pr+qr)^2-3(p+q+r)pqr\Bigr]$$

will be positive for general distinct $p,q,r$). However, if by chance $p+q+r=0$, the quadratic reduces to a linear equation

$$2(pq+pr+qr)x+3pqr=0,$$

giving exactly one real root. So, the answer is:

• Two real roots (if $p+q+r\ne0$);
• One real root (if $p+q+r=0$).

Answer to (4):
When $p,q,r \in \mathbb{R}^+$ (i.e. all are positive), the coefficients of the quadratic are all positive. In fact,

• The coefficient of $x^2$ is $p+q+r>0$,
• The coefficient of $x$ is $2(pq+pr+qr)>0$, and
• The constant term is $3pqr>0$.

Thus, for $x>0$ every term is positive so the left‐side is positive and the quadratic cannot vanish for any $x>0$. (Moreover, from the Vieta’s formula the sum of the roots is negative and their product is positive so both roots are negative.)

Hence, there are no positive real roots.