If the roots of the equation frac{1}{x + p} + | vedclass

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

Question

(A) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ Multiply both sides by $r(x + p)(x + q)$ to clear the denominators: $r(x + q

Vedclass · Accepted Answer

$\frac{p + q}{2}$

Vedclass · Answer

(A) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ Multiply both sides by $r(x + p)(x + q)$ to clear the denominators: $r(x + q) + r(x + p) = (x + p)(x + q)$ $rx + rq + rx + rp = x^2 + px + qx + pq$ $2rx + r(p + q) = x^2 + (p + q)x + pq$ Rearranging into standard quadratic form $ax^2 + bx + c = 0$: $x^2 + (p + q - 2r)x + (pq - r(p + q)) = 0$ Let the roots be $\alpha$ and $-\alpha$. For a quadratic equation $x^2 + Bx + C = 0$,the sum of roots is $-B$. Here,$\alpha + (-\alpha) = 0$,so the coefficient of $x$ must be $0$. $p + q - 2r = 0$ $2r = p + q$ $r = \frac{p + q}{2}$

If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign,then $r = ......$

Similar Questions

The value of $4+\frac{1}{5+\frac{1}{4+\frac{1}{5+\frac{1}{4+\ldots \ldots \infty}}}}$ is

If $a, b$ and $c$ are positive numbers in a $GP$, then the roots of the quadratic equation $(\log_{e} a) x^{2} - (2 \log_{e} b) x + (\log_{e} c) = 0$ are

If $b_{1} b_{2} = 2(c_{1} + c_{2})$ and $b_{1}, b_{2}, c_{1}, c_{2}$ are all real numbers, then at least one of the equations $x^{2} + b_{1} x + c_{1} = 0$ and $x^{2} + b_{2} x + c_{2} = 0$ has

If the roots of the given equation $(\cos p-1) x^2+(\cos p) x+\sin p=0$ are real,then

$\alpha, \beta, \gamma, 2, \varepsilon$ are the roots of the equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$. If $\alpha < \beta < \gamma < 2 < \varepsilon$,then find the value of $\alpha+2 \beta+3 \gamma+5 \varepsilon$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module