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) Given the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + rq +

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.$rx + rq + rx + rp = x^2 + px + qx + pq$.$2rx + r(p+q) = x^2 + (p+q)x + pq$.Rearranging into standard quadratic form $ax^2 + bx + c = 0$:$x^2 + (p+q-2r)x + (pq - rp - rq) = 0$.Let the roots be $\alpha$ and $-\alpha$ since 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 $p+q-2r = 0$,so $2r = p+q$.The product of the roots is $\alpha 	imes (-\alpha) = -\alpha^2$.From the equation,the product of roots is $pq - r(p+q)$.Substituting $r = \frac{p+q}{2}$ into the product:Product $= pq - (\frac{p+q}{2})(p+q) = pq - \frac{(p+q)^2}{2}$.Product $= \frac{2pq - (p^2 + 2pq + q^2)}{2} = \frac{2pq - p^2 - 2pq - q^2}{2} = -\frac{p^2 + q^2}{2}$.

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,what is their product?

Similar Questions

If $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$,then $x^2(x-4)^2$ is equal to :

If $p$ and $q$ are prime numbers satisfying the condition $p^{2}-2q^{2}=1$,then the value of $p^{2}+2q^{2}$ is

If $\sinh(\log x) = -2$,then $x =$

If the roots of the equation $3x^2 + 4kx + 3 = 0$ are non-real,then $k$ lies in the interval

If $ax^2 + bx + c = 0$ has real and distinct roots,$\alpha$ and $\beta$ where $\beta > \alpha$. Further,if $a > 0, b < 0$,and $c < 0$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module