જો p, q અને r (p ne q,r ne 0), વાસ્તવિક સંખ્ય | vedclass

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

Question

$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$

$\frac{{x + p + x + q}}{{(x + p)(x + q)}} = \frac{1}{r}$

$(2x + p + q)r = {x^2} + px + qx + pq$

${x

Vedclass · Accepted Answer

${p^2} + {q^2}$

Vedclass · Answer

$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$

$\frac{{x + p + x + q}}{{(x + p)(x + q)}} = \frac{1}{r}$

$(2x + p + q)r = {x^2} + px + qx + pq$

${x^2} + (p + q - 2r)x + pq - pr - qr = 0$

Let $\alpha$ and $\beta$ be the roots.

$	herefore \alpha+\beta=-(p+q-2 r)..........(i)$

$\alpha \beta=pq-pr-qr..........(ii)$

$\because \alpha=-\beta(	ext { given })$

$	herefore$ in eq. $(1),$ we get

$\Rightarrow \quad-(p+q-2 r)=0..........(iii)$

Now, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$

$=(-(p+q-2 r))^{2}-2(p q-p r-q r) \ldots$ (from $(i)$ and $(ii))$

$=p^{2}+q^{2}+4 r^{2}+2 p q-4 p r-4 q r-2 p q+2 p r+2 q r$

$=p^{2}+q^{2}+4 r^{2}-2 p r-2 q r$

$=p^{2}+q^{2}+2 r(2 r-p-q) \ldots(	ext { from (iii) })$

$=p^{2}+q^{2}+0$

$=p^{2}+q^{2}$

Similar Questions