Question

Find the roots of the equation:
$\dfrac{1}{{x + 4}} - \dfrac{1}{{x - 7}} = \dfrac{{11}}{{30}}\,\,\,x \ne - 4,7$

Answer 1

Hint: Simplify the equation you will get the quadratic equation and solve that equation you will get answers and also roots cannot be equal to $ - 4\,\,or\,\,7$ because at $x = - 4\,\,\,\& \,\,x = 7$, the equation becomes undefined as denominator becomes zero.

Complete step-by-step answer:
So here we are given with an equation
\[\dfrac{1}{{x + 4}} - \dfrac{1}{{x - 7}} = \dfrac{{11}}{{30}}\,\,\,\]where \[x \ne - 4,7\]because if $x = - 4\,\,\,\& \,\,x = 7$, the equation becomes undefined as denominator becomes zero.
Now on solving we get
\[
   \Rightarrow \dfrac{{\left( {x - 7} \right) - \left( {x + 4} \right)}}{{\left( {x + 4} \right)\left( {x - 7} \right)}} = \dfrac{{11}}{{30}} \\
  {\text{taking LCM both sides we get}} \\
   \Rightarrow \dfrac{{x - 7 - x - 4}}{{\left( {{x^2} + 4x - 7x - 28} \right)}} = \dfrac{{11}}{{30}} \\
   \Rightarrow \dfrac{{ - 11}}{{\left( {{x^2} - 3x - 28} \right)}} = \dfrac{{11}}{{30}} \\
  {\text{upon cross multiplication we get}} \\
   \Rightarrow \dfrac{{ - 1}}{{\left( {{x^2} - 3x - 28} \right)}} = \dfrac{1}{{30}} \\
   \Rightarrow - 30 = {x^2} - 3x - 28 \\
   \Rightarrow {x^2} - 3x - 28 + 30 = 0 \\
   \Rightarrow {x^2} - 3x + 2 = 0 \\
 \]
Here we got the quadratic, now it has power \[2\], so that \[2\]roots are possible. So let $\alpha \,\,and\,\,\beta $are the roots of this equation and we know if any equation is in form of $a{x^2} + bx + c = 0$
Then roots,
$\alpha = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $\beta = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Here we have equation \[{x^2} - 3x + 2 = 0\]
So here $a = 1\,,\,b = - 3\,,\,c = 2\,$
$
  \alpha = \dfrac{{ + 3 + \sqrt {{{( - 3)}^2} - 4 \times 1 \times 2} }}{{2 \times 1}}\,\,and\,\,\beta = \dfrac{{ + 3 - \sqrt {{{( - 3)}^2} - 4 \times 1 \times 2} }}{{2 \times 1}} \\
  \alpha = \dfrac{{ + 3 + \sqrt {9 - 8} }}{2}\,\,and\,\,\beta = \dfrac{{ + 3 - \sqrt {9 - 8} }}{2} \\
  \alpha = \dfrac{{ + 3 + 1}}{2}\,\,and\,\,\beta = \dfrac{{ + 3 - 1}}{2} \\
  \alpha = 2\,\,and\,\,\beta = 1 \\
 $

So here $\left( {\alpha ,\beta } \right) = \left( {2,1} \right)$ are the roots of this equation.

Note: Here we when simplify we will get ${x^2} - 3x + 2 = 0$this we can also by this method:
$
  {x^2} - 3x + 2 = 0 \\
  {x^2} - 2x - x + 2 = 0 \\
  x(x - 2) - 1(x - 2) = 0 \\
  (x - 1)(x - 2) = 0 \\
  x - 2 = 0\,\,\& \,\,x - 1 = 0 \\
  x = 2\,\,\& \,\,x = 1 \\
 $
So $2\,\,\& \,\,1$ are the roots of this equation.