Question

simplify: p/px-1 + q/qx-1 = p+q ​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\bf :\longmapsto\:Simplify : \: \dfrac{p}{px - 1} + \dfrac{q}{qx - 1} = p + q$

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

$\rm :\longmapsto\:\dfrac{p}{px - 1} + \dfrac{q}{qx - 1} = p + q$

$\rm :\longmapsto\:\dfrac{p}{px - 1} + \dfrac{q}{qx - 1} - p - q = 0$

$\rm :\longmapsto\:\bigg(\dfrac{p}{px - 1} - q\bigg) + \bigg(\dfrac{q}{qx - 1} - p\bigg) = 0$

$\rm :\longmapsto\:\bigg(\dfrac{p - q(px - 1)}{px - 1}\bigg) + \bigg(\dfrac{q - p(qx - 1)}{qx - 1}\bigg) = 0$

$\rm :\longmapsto\:\bigg(\dfrac{p - qpx + q}{px - 1}\bigg) + \bigg(\dfrac{q - pqx + p}{qx - 1}\bigg) = 0$

$\rm :\longmapsto\:(q + p - pqx)\bigg(\dfrac{1}{px - 1}+\dfrac{1}{qx - 1}\bigg) = 0$

$\rm :\longmapsto\:(q + p - pqx)\bigg(\dfrac{qx - 1 + px - 1}{(px - 1)(qx - 1)}\bigg) = 0$

$\rm :\longmapsto\:(q + p - pqx)\bigg(\dfrac{qx+ px - 2}{(px - 1)(qx - 1)}\bigg) = 0$

$\rm :\implies\:p + q - pqx = 0 \: \: or \: px + qx - 2 = 0$

$\rm :\implies\:p + q = pqx \: \: or \: x(p + q)= 2$

$\bf\implies \:x = \dfrac{p + q}{pq} \: \: \: \: \: or \: \: \: \: \: x = \dfrac{2}{p + q}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac