Question

The solution set of the equation, $ \dfrac{x}{x-4}-\dfrac{1}{x+3}=\dfrac{28}{{{x}^{2}}-x-12} $ is,
[a]– 4
[b] 4, -6
[c] – 6
[d] 4

Answer 1

Hint: Do the cross multiplication in the given equation and simplify until you get a quadratic equation. Then solve the quadratic equation by any method to get the final answer.

Complete step-by-step answer:
To find the solution of the given question, we will first write down the given equation, therefore,
 $ \dfrac{x}{x-4}-\dfrac{1}{x+3}=\dfrac{28}{{{x}^{2}}-x-12} $
If we do the cross multiplication in the right-hand side of the above equation, we will get,
\[\therefore \dfrac{x\left( x+3 \right)-\left( x-4 \right)}{\left( x-4 \right)\left( x+3 \right)}=\dfrac{28}{{{x}^{2}}-x-12}\]
If we multiply inside the bracket of the above equation, we will get,
\[\therefore \dfrac{{{x}^{2}}+3x-x+4}{\left( x-4 \right)\left( x+3 \right)}=\dfrac{28}{{{x}^{2}}-x-12}\]
Now, we know that
 $ \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab $
Hence, we have
 $ \left( x-4 \right)\left( x+3 \right)={{x}^{2}}-x-12 $
Hence the above equation becomes
 $ \begin{align}
  & \dfrac{{{x}^{2}}+2x+4}{{{x}^{2}}-x-12}=\dfrac{28}{{{x}^{2}}-x-12} \\
 & \Rightarrow {{x}^{2}}+2x+4=28 \\
\end{align} $
If we shift 28 on the left-hand side of the equation, we will get,
\[\therefore {{x}^{2}}+2x+4-28=0\]
If we simplify the above equation, we will get,
\[\therefore {{x}^{2}}+2x-24=0\]
In the above quadratic equation we can replace 2x by 6x – 4x as 6x – 4x = 2x and \[6\times \left( -4 \right)=-24\]therefore we will get,
\[\therefore {{x}^{2}}+6x-4x-24=0\]
If we take ‘x’ common from the first two terms of the above equation, we will get,
\[\therefore \left( x+6 \right)x-4x-24=0\]
Also by taking -4 common from the last two terms of the above equation, we will get,
\[\therefore \left( x+6 \right)x-4\left( x+6 \right)=0\]
If we take (x + 6) common from the above equation, we will get,
\[\therefore \left( x+6 \right)\left( x-4 \right)=0\]
Above equation can also be written as,
x + 6 = 0 OR x – 4 = 0
Therefore, x = - 6 OR x = 4
Therefore the solution set of the given equation $ \dfrac{x}{x-4}-\dfrac{1}{x+3}=\dfrac{28}{{{x}^{2}}-x-12} $ is 4, - 6.
Therefore the correct answer is option (b).

Note: You can solve the quadratic equation \[{{x}^{2}}+2x-24=0\] by using the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] as it will give you answer quickly.
Here a = 1, b=2 and c= -24
Hence, we have
 $ x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( 1 \right)\left( -24 \right)}}{2}=\dfrac{-2\pm \sqrt{4+96}}{2}=\dfrac{-2\pm 10}{2} $
Taking the positive sign, we get
 $ x=\dfrac{-2+10}{2}=4 $
Taking the negative sign, we get
 $ x=\dfrac{-2-10}{2}=-6 $
Hence, we have x= 4 or x = -6
Therefore the correct answer is option (b).