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

(B) The given equation is $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + r

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given equation is $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + rq + rx + rp = x^2 + px + qx + pq$.Rearranging the terms to form a quadratic equation: $x^2 + x(p + q - 2r) + (pq - pr - qr) = 0$.Let the roots be $\alpha$ and $-\alpha$ as they are equal in magnitude but opposite in sign.The sum of the roots is $\alpha + (-\alpha) = 0$.From the quadratic equation,the sum of roots is $-(p + q - 2r) = 0$,which implies $p + q - 2r = 0$,or $r = \frac{p + q}{2}$.The product of the roots is $\alpha \cdot (-\alpha) = -\alpha^2$.From the quadratic equation,the product of roots is $pq - r(p + q)$.Substituting $r = \frac{p + q}{2}$ into the product expression:Product $= pq - \frac{p + q}{2}(p + q) = pq - \frac{(p + q)^2}{2} = \frac{2pq - (p^2 + 2pq + q^2)}{2} = \frac{2pq - p^2 - 2pq - q^2}{2} = -\frac{p^2 + q^2}{2}$.

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 the product of the roots will be

Similar Questions

Let one root of $ax^2 + bx + c = 0$,where $a, b, c$ are integers,be $3 + \sqrt{5}$. Then the other root is:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-16x+63=0$
$II.$ $y^{2}-2y-35=0$

The roots of the quadratic equation $(a + b - 2c)x^2 - (2a - b - c)x + (a - 2b + c) = 0$ are

Let $y = \sqrt {\frac{{(x + 1)(x - 3)}}{{(x - 2)}}} $,then all real values of $x$ for which $y$ takes real values,are

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2} = 14641$
$II.$ $y = \sqrt{14641}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module