Question

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

Answer 1

Hint: To solve the given question, we will first take LCM on the left-hand side of the equation given in the question. After taking the LCM, we will multiply both sides of the equation by (x + 4) (x – 7). After simplifying the equation, we will get a quadratic equation in x. To find the roots of this quadratic equation, we will use the completing square method. In this method, we will try to make square terms on both sides. After that, we will take the square root on both sides to get the required roots of x.

Complete step-by-step answer:
To start with, we have the following equation given in the question.
\[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30}\]
Now, we will take LCM on the left-hand side of the above equation. Thus, we will get the following equation.
\[\Rightarrow \dfrac{\left( x-7 \right)-\left( x+4 \right)}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\]
\[\Rightarrow \dfrac{x-7-x-4}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\]
\[\Rightarrow \dfrac{-11}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}......\left( i \right)\]
Now, we will multiply both sides of the equation by (x + 4) (x – 7). Thus, we will get the following equation.
\[\Rightarrow \dfrac{-11}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\left( x+4 \right)\left( x-7 \right)\]
\[\Rightarrow -11=\dfrac{11}{30}\left( x+4 \right)\left( x-7 \right)\]
Now, we will divide both sides of the equation by 11. Thus, we will get the following equation.
\[\Rightarrow -1=\dfrac{1}{30}\left( x+4 \right)\left( x-7 \right)\]
On cross multiplying 30, we will get,
\[\Rightarrow -30=\left( x+4 \right)\left( x-7 \right)\]
\[\Rightarrow \left( x+4 \right)\left( x-7 \right)+30=0\]
\[\Rightarrow {{x}^{2}}+4x-7x-28+30=0\]
\[\Rightarrow {{x}^{2}}-3x+2=0\]
The above equation is a quadratic equation in x. A quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] can be solved with the help of completing the square method. In this method, we will try to make the square terms on both sides of the equation. Thus, we have,
\[\Rightarrow {{x}^{2}}-3x+2=0\]
\[\Rightarrow {{x}^{2}}-2\times \left( \dfrac{3}{2} \right)\times x+{{\left( \dfrac{3}{2} \right)}^{2}}+2-{{\left( \dfrac{3}{2} \right)}^{2}}=0\]
\[\Rightarrow {{x}^{2}}-2\times \left( \dfrac{3}{2}\times x \right)+{{\left( \dfrac{3}{2} \right)}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}}-2\]
Now, we will sue the identity, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.\] Thus, we will get,
\[{{\left( x-\dfrac{3}{2} \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\]
On taking the square root on both the sides, we will get,
\[\Rightarrow \left( x-\dfrac{3}{2} \right)=\pm \left( \dfrac{1}{2} \right)\]
\[\Rightarrow \left( x-\dfrac{3}{2} \right)=\left( \dfrac{1}{2} \right)\text{ or }\left( x-\dfrac{3}{2} \right)=\left( \dfrac{-1}{2} \right)\]
\[\Rightarrow x=\dfrac{3}{2}+\dfrac{1}{2}\text{ or }x=\dfrac{3}{2}-\dfrac{1}{2}\]
\[\Rightarrow x=\dfrac{4}{2}\text{ or }x=\dfrac{2}{2}\]
\[\Rightarrow x=2\text{ or }x=1\]
Thus, the values of x which satisfy the equation given in the question are x = 2 and x = 1.
Thus, the roots of the equation are x = 1, 2.

Note: The quadratic equation obtained in the question can also be solved by the factorization method. In the factorization method, we write \[{{x}^{2}}+ax+b=0\] in the form \[\left( x-\alpha \right)\left( x-\beta \right)=0,\] where \[\alpha \] and \[\beta \] are the roots of the equation. Thus, we have,
\[{{x}^{2}}-3x+2=0\]
\[\Rightarrow {{x}^{2}}-x-2x+2=0\]
\[\Rightarrow x\left( x-1 \right)-2\left( x-1 \right)=0\]
\[\Rightarrow \left( x-2 \right)\left( x-1 \right)=0\]
Thus, the roots of the equation are x = 2 and x = 1.