Question

if the roots of 1/x-p+1/x-q=1/r are equal in magnitude but opposite in sign, then find the product of roots in p and q only​

Answer 1

Step-by-step explanation:

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Answer

Correct option is

C

−(

2

p

2

+q

2

)= Product of roots

x+p

1

+

x+q

1

=

r

1

⇒(x+q+x+p)r=(x+p)(x+q)

⇒2xr+(p+q)r=x

2

+(p+q)x+pq

⇒x

2

+(p+q−2r)x+pq−pr−qr=0

As roots are equal in magnitude and opposite the sign

Then sum of roots =0

⇒p+q=2r

Hence, product of roots

pq−pr−qr=pq−

2

(p+q)

2

=−

2

p

2

+q

2

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Answer 2

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\:\dfrac{1}{x - p} + \dfrac{1}{x - q} = \dfrac{1}{r}$

$\rm :\longmapsto\:\dfrac{x - q + x - p}{(x - p)(x - q)} = \dfrac{1}{r}$

$\rm :\longmapsto\:\dfrac{2x - q - p}{(x - p)(x - q)} = \dfrac{1}{r}$

$\rm :\longmapsto\:(x - p)(x - q) = r(2x - p - q)$

$\rm :\longmapsto\: {x}^{2} - px - qx + pq = 2rx - pr - qr$

$\rm :\longmapsto\: {x}^{2} - px - qx + pq - 2rx + pr + qr = 0$

$\rm :\longmapsto\: {x}^{2} - px - qx - 2rx + pq + pr + qr = 0$

$\rm :\longmapsto\: {x}^{2} - x(p + q + 2r) + (pq + pr + qr) = 0$

So, its a quadratic equation.

Now,

$\red{\rm :\longmapsto\:\:Let \: \alpha , \: \beta \: be \: the \: roots \: of \: above \: quadratic.}$

Further, It is given that,

$\rm :\longmapsto\: \beta = - \: \alpha$

Now, we know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha + \beta = - \: \dfrac{ - (p + q + 2r)}{1}$

$\rm :\longmapsto\: \alpha - \alpha = p + q + 2r$

$\rm :\longmapsto\: 0 = p + q + 2r$

$\bf\implies \:r = - \: \dfrac{p + q}{2}$

Now,

We know that,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\: \alpha \beta = \dfrac{pq + qr + rp}{1}$

$\rm :\longmapsto\: \alpha \beta = pq + r(q + p)$

$\rm :\longmapsto\: \alpha \beta = pq - \dfrac{(p + q)}{2} (q + p)$

$\rm :\longmapsto\: \alpha \beta = pq - \dfrac{(p + q) {}^{2} }{2}$

$\rm :\longmapsto\: \alpha \beta = pq - \dfrac{ {p}^{2} + {q}^{2} + 2pq }{2}$

$\rm :\longmapsto\: \alpha \beta = \dfrac{2pq - {p}^{2} - {q}^{2} - 2pq }{2}$

$\rm :\longmapsto\: \alpha \beta = \dfrac{ -{p}^{2}-{q}^{2}}{2}$

$\rm :\longmapsto\: \alpha \beta = - \: \dfrac{{p}^{2} + {q}^{2}}{2}$

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$