Given that $r$ and $s$ are constants , the solution of the quadratic equation, $r{x}^{2}=\dfrac{1}{s}x+3$ , is:
- $x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } +12r } }{ 2r } $
- $x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { -\dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2sr } $
- $x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2r } $
- $x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { { s }^{ 2 }-12sr } }{ 2sr } $
The equation is rx^2 - (1/s)x - 3 = 0. Using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)] / 2a, where a=r, b=-1/s, c=-3. x = [1/s +/- sqrt(1/s^2 - 4*r*(-3))] / 2r = [1/s +/- sqrt(1/s^2 + 12r)] / 2r. This simplifies to 1/(2sr) +/- sqrt(1/s^2 + 12r) / 2r.
We first rewrite the equation into the standard quadratic form rsx^2 - x - 3s = 0. Using the quadratic formula, x equals [-(-1) plus or minus the square root of (-1)^2 - 4(rs)(-3s)] divided by 2rs, which simplifies to [1 plus or minus the square root of (1 + 12rs^2)] divided by 2rs. By separating the denominator for the first term, this is equivalent to x = 1/(2sr) plus or minus the square root of [1/s^2 + 12r] divided by 2r.