Question

Using factorization, find the roots of the quadratic equation $\dfrac{1}{x-3}-\dfrac{1}{x+5}=\dfrac{1}{6},x\ne 3,-5$ .

Answer 1

Hint: Firstly, we have to simplify the given expression using the rules $\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd}$ and cross multiplication method( if $\dfrac{a}{b}=\dfrac{c}{d}$ then $ad=bc$ ). Then, we have to simplify the equation and make it in the form of the quadratic equation $a{{x}^{2}}+bx+c=0$ . Finally, we have to factorize the equation by either splitting the middle terms or quadratic formula or any other method and find the values of x.

Complete step by step answer:
We have to find the roots of the quadratic equation $\dfrac{1}{x-3}-\dfrac{1}{x+5}=\dfrac{1}{6}$ by factorization. We know that $\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd}$ . Therefore, we can write the given expression as
$\begin{align}
  & \Rightarrow \dfrac{x+5-\left( x-3 \right)}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{6} \\
 & \Rightarrow \dfrac{x+5-x+3}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{6} \\
\end{align}$
Let us add the like terms in the numerator of the LHS.
$\begin{align}
  & \Rightarrow \dfrac{\require{cancel}\cancel{x}+5\require{cancel}\cancel{-x}+3}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{6} \\
 & \Rightarrow \dfrac{5+3}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{6} \\
 & \Rightarrow \dfrac{8}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{6} \\
\end{align}$
Now, we have to cross multiply, that is, if $\dfrac{a}{b}=\dfrac{c}{d}$ then $ad=bc$ .
$\begin{align}
  & \Rightarrow 8\times 6=\left( x-3 \right)\left( x+5 \right) \\
 & \Rightarrow 48=\left( x-3 \right)\left( x+5 \right) \\
\end{align}$
We know that $\left( x+a \right)\left( x-b \right)={{x}^{2}}+\left( a-b \right)x-ab$ . Therefore, we can write the RHS of the above equation as
$\begin{align}
  & \Rightarrow 48=\left( x+5 \right)\left( x-3 \right) \\
 & \Rightarrow 48={{x}^{2}}+\left( 5-3 \right)x-5\times 3 \\
 & \Rightarrow 48={{x}^{2}}+2x-15 \\
\end{align}$
Let us move 48 to the RHS.
$\begin{align}
  & \Rightarrow {{x}^{2}}+2x-15-48=0 \\
 & \Rightarrow {{x}^{2}}+2x-63=0 \\
\end{align}$
We have to factorize the above equation. Let us do this by splitting the middle term. We can see that the sum is 2 and the product is -63. We know that $-7\times 9=-63$ and $-7+9=2$ . Therefore, we can factorize the above expression into
$\Rightarrow {{x}^{2}}+9x-7x-63=0$
Let us take the common terms outside.
$\Rightarrow x\left( x+9 \right)-7\left( x+9 \right)=0$
We have to take $x+9$ outside.
$\Rightarrow \left( x+9 \right)\left( x-7 \right)=0$
From the above equation, we can conclude that $\left( x+9 \right)=0,\left( x-7 \right)=0$ . Let us consider $\left( x+9 \right)=0$ . We have to collect the constants on one side by moving 9 to the RHS.
$\Rightarrow x=-9$
Now, let us consider $x-7=0$ . We have to move -7 to the RHS.
$\Rightarrow x=7$
Therefore, the roots of the quadratic equation $\dfrac{1}{x-3}-\dfrac{1}{x+5}=\dfrac{1}{6}$ is -9 and 7.

Note: Students must be very thorough with the properties, identities and rules of the algebra. They must clearly study the methods to factorize polynomials. In the above solution, we have factorized the equation by splitting the middle term. Students can also do this by quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for a quadratic equation $a{{x}^{2}}+bx+c=0$ . This method will be time consuming.