Question

Solve
 $ \dfrac{2}{{x + 1}} + \dfrac{3}{{2\left( {x - 2} \right)}} = \dfrac{{23}}{{5x}},\,\,x \ne 0, - 1,2 $

Answer 1

Hint: For this type of problems we first simplify both side by taking L.C.M. of fraction terms present on both side and then cross multiplying to form either a linear equation or a quadratic equation and then solving an equation so obtained to find value of ‘x’ or require solution of given problem.

Complete step-by-step answer:
Given equation is $ \dfrac{2}{{x + 1}} + \dfrac{3}{{2\left( {x - 2} \right)}} = \dfrac{{23}}{{5x}},\,\,x \ne 0, - 1,2 $
To solve the above equation. We first take L.C.M. of left hand side term and then cross multiplying to form an equation.
\[
  \dfrac{{4\left( {x - 2} \right) + 3\left( {x + 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 2} \right)}} = \dfrac{{23}}{{5x}} \\
   \Rightarrow \dfrac{{4x - 8 + 3x + 3}}{{2\left( {x + 1} \right)\left( {x - 2} \right)}} = \dfrac{{23}}{{5x}} \\
   \Rightarrow \dfrac{{7x - 5}}{{2\left( {x + 1} \right)\left( {x - 2} \right)}} = \dfrac{{23}}{{5x}} \\
 \]
Cross multiplying above. We have,
 $
  5x\left( {7x - 5} \right) = 23\left\{ {2\left( {x + 1} \right)\left( {x - 2} \right)} \right\} \\
   \Rightarrow 35{x^2} - 25x = 46\left( {{x^2} - 2x + x - 2} \right) \\
   \Rightarrow 35{x^2} - 25x = 46\left( {{x^2} - x - 2} \right) \\
   \Rightarrow 35{x^2} - 25x = 46{x^2} - 46x - 92 \\
   \Rightarrow 35{x^2} - 25x - 46{x^2} + 46x + 92 = 0 \\
   \Rightarrow - 11{x^2} + 21x + 92 = 0 \\
  or \\
  11{x^2} - 21x - 92 = 0 \;
  $
Now, solving the above equation by quadratic formula method.
Comparing the above formed equation with standard quadratic equation $ a{x^2} + bx + c = 0 $ . We have,
 $ a = 11,\,\,b = - 21\,\,and\,\,c = - 92 $
Calculating discriminant of quadratic. Given as:
 $ D = {b^2} - 4ac $
Substituting values of a, b and c in above.
 $
  D = {\left( { - 21} \right)^2} - 4\left( {11} \right)\left( { - 92} \right) \\
   \Rightarrow D = 441 + 4048 \\
   \Rightarrow D = 4489 \;
  $
As, $ D > 0 $ Therefore, roots of a quadratic equations are real but unequal and given as:
 $ \alpha = \dfrac{{ - b + \sqrt D }}{{2a}},\,\,\beta = \dfrac{{ - b - \sqrt D }}{{2a}} $
Substituting values in above. We have,
 $
  \alpha = \dfrac{{ - \left( { - 21} \right) + \sqrt {4489} }}{{2(11)}},\,\,\beta = \dfrac{{ - \left( { - 21} \right) - \sqrt {4489} }}{{2(11)}} \\
   \Rightarrow \alpha = \dfrac{{21 + 67}}{{22}},\,\,\beta = \dfrac{{21 - 67}}{{22}} \\
   \Rightarrow \alpha = \dfrac{{88}}{{22}},\,\,\beta = \dfrac{{ - 46}}{{22}} \\
   \Rightarrow \alpha = 4,\,\,\beta = - \dfrac{{23}}{{11}} \\
  $
Or we can say that roots of quadratic equation are $ x = 4\,\,and\,\,x = - \dfrac{{23}}{{11}} $
Therefore, from above we can see that solution of given equation $ \dfrac{2}{{x + 1}} + \dfrac{3}{{2\left( {x - 2} \right)}} = \dfrac{{23}}{{5x}},\,\,x \ne 0, - 1,2 $ are $ x = 4\,\,and\,\,x = - \dfrac{{23}}{{11}} $
So, the correct answer is “$ x = 4\,\,and\,\,x = - \dfrac{{23}}{{11}} $ ”.

Note: In this type of problems one can solve by first shifting all terms to the left side and then simplifying it by taking LCM and solving the equation so formed. If the equation so form will be linear then one directly finds its solution but if it will be a quadratic then there different ways to solve a quadratic equation like middle term splitting method, completing square method and quadratic formula method. One can apply either one as the answer will be the same in any case.