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

Let $a$ be an integer selected at random from the set $\{0, 1, 2, 3, ..., 9\}$. The probability that the equation $ax^2 - ax + 1 = 0$ has real roots is ...

If $-1+i$ is a root of the equation $x^4+4x^3+5x^2+2x-2=0$,then the real roots of this equation are

The equation $\sqrt{3x^2 + x + 5} = x - 3$,where $x$ is real,has

If the values of $k$ for which the equation $x^2+2(k+2)x+6k+7=0$ has equal roots are $k_1$ and $k_2$,then $k_1^2+k_2^2=$

If the expression $x^3+3x^2-9x+\lambda$ is of the form $(x-\alpha)^2(x-\beta)$,then the values of $\lambda$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module