The solution of the equation frac{p + q - x}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equation: $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$Add $3$ to both sides of the equation:

Vedclass · Accepted Answer

$x = p + q + r$

Vedclass · Answer

(A) Given equation: $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$Add $3$ to both sides of the equation:$(\frac{p + q - x}{r} + 1) + (\frac{q + r - x}{p} + 1) + (\frac{r + p - x}{q} + 1) + \frac{3x}{p + q + r} = 3$$\frac{p + q + r - x}{r} + \frac{q + r + p - x}{p} + \frac{r + p + q - x}{q} + \frac{3x}{p + q + r} = 3$$(p + q + r - x) (\frac{1}{r} + \frac{1}{p} + \frac{1}{q}) = 3 - \frac{3x}{p + q + r}$$(p + q + r - x) (\frac{1}{p} + \frac{1}{q} + \frac{1}{r}) = 3 (\frac{p + q + r - x}{p + q + r})$$(p + q + r - x) [\frac{1}{p} + \frac{1}{q} + \frac{1}{r} - \frac{3}{p + q + r}] = 0$This implies $p + q + r - x = 0$ or the bracket term is zero.Therefore,$x = p + q + r$.

The solution of the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$ is

Similar Questions

The roots of the equation $|x^2-x-6|=x+2$ are

$A$ two-digit number is four times the sum of its digits and three times the product of its digits. The number is:

The number of roots of the equation $|x|^2 - 7|x| + 12 = 0$ is

The smallest value of the constant $m > 0$ for which $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$ is:

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ for $x \in R$ has:

Vedclass Test Series

Exam Paper Generator

Online Exam Module