Question

Solve the following equation:
\[\dfrac{24}{18-x}-\dfrac{24}{18+x}=1\]

Answer 1

Hint: We solve this problem by adding the terms using the LCM method so that we get a quadratic equation in \[x\]
Then we use the formula of roots of quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We also use the formula that the difference of square of two numbers that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]

Complete step by step answer:
We are given that the equation to be solved as
\[\dfrac{24}{18-x}-\dfrac{24}{18+x}=1\]
Now, by adding the terms in the above equation using the LCM method we get
\[\begin{align}
  & \Rightarrow \dfrac{24\left( 18+x \right)-24\left( 18-x \right)}{\left( 18+x \right)\left( 18-x \right)}=1 \\
 & \Rightarrow \dfrac{432+24x-432+24x}{\left( 18+x \right)\left( 18-x \right)}=1 \\
 & \Rightarrow \dfrac{48x}{\left( 18+x \right)\left( 18-x \right)}=1 \\
\end{align}\]
Now, by cross multiplying the terms in the above equation we get
\[\Rightarrow 48x=\left( 18+x \right)\left( 18-x \right)\]
We know that the formula that the difference of square of two numbers that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow 48x={{18}^{2}}-{{x}^{2}} \\
 & \Rightarrow {{x}^{2}}+48x-324=0 \\
\end{align}\]
We know that the formula of roots of quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By using this formula of roots of quadratic equation to above equation we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-48\pm \sqrt{{{\left( -48 \right)}^{2}}-4\left( 1 \right)\left( -324 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-48\pm \sqrt{2304+1296}}{2} \\
 & \Rightarrow x=\dfrac{-48\pm \sqrt{3600}}{2} \\
\end{align}\]
Here, we can see that the number 3600 is perfect square that is \[\sqrt{3600}=60\]
By substituting this value in above equation we get
\[\Rightarrow x=\dfrac{-48\pm 60}{2}\]
Here, we can see that we get two values of \[x\] one for positive and one for negative sig.
By taking the positive sign in above equation we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-48+60}{2} \\
 & \Rightarrow x=\dfrac{12}{2}=6 \\
\end{align}\]
Now, by taking the negative sign we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-48-60}{2} \\
 & \Rightarrow x=\dfrac{-108}{2}=-54 \\
\end{align}\]
Therefore we can conclude that the roots of the given equation are 6 and -54.

Note:
 We can solve the quadratic equation in another method.
We have the quadratic equation as
\[\Rightarrow {{x}^{2}}+48x-324=0\]
Now, let us use the factorisation method that is by dividing the number 48 into two numbers such that the product will be -324 then we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+54x-6x-324=0 \\
 & \Rightarrow x\left( x+54 \right)-6\left( x+54 \right)=0 \\
 & \Rightarrow \left( x+54 \right)\left( x-6 \right)=0 \\
\end{align}\]
We know that if \[a\times b=0\] then either of \[a,b\] is zero.
By taking the first term to 0 we get
\[\begin{align}
  & \Rightarrow x+54=0 \\
 & \Rightarrow x=-54 \\
\end{align}\]
Now, by taking the second term to 0 we get
\[\begin{align}
  & \Rightarrow x-6=0 \\
 & \Rightarrow x=6 \\
\end{align}\]
Therefore we can conclude that the roots of given equation are 6 and -54