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

Let $r(x)$ be the remainder when the polynomial $x^{135}+x^{125}-x^{115}+x^5+1$ is divided by $x^3-x$. Then,

If the equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$ has equal roots,where $a + c = 15$ and $b = \frac{36}{5}$,then $a^2 + c^2$ is equal to . . . . . .

The roots of the cubic equation $3x^3+4x^2-5x-2=0$ are diminished by $h$,and a cubic equation with these diminished roots is formed. If the transformed equation does not contain the $x^2$ term,then the roots of the transformed equation are

Let $\theta$ be an acute angle such that the equation $x^3+4 x^2 \cos \theta+x \cot \theta=0$ has multiple roots. Then the value of $\theta$ (in radians) is

The number of solution$(s)$ of the equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ is/are

Vedclass Test Series

Exam Paper Generator

Online Exam Module