Question

Solve the following quadratic equation by factorisation method:
\[\dfrac{x-2}{x-3}+\dfrac{x-4}{x-5}=\dfrac{10}{3}\] where \[x\ne 3,5\]
The roots are \[6,\dfrac{7}{2}\]
(a) True
(b) False

Answer 1

Hint: We solve this problem first by converting the given equation into a quadratic equation of the form
\[a{{x}^{2}}+bx+c=0\]
Then we use the factorisation method that is we take the middle term coefficient \['b'\] as sum of two numbers \[p,q\] such that \[p\times q=a\times c\]
By writing in that form we get the factors which will be easy to solve the equation.

Complete step-by-step answer:
We are given that the equation as
\[\Rightarrow \dfrac{x-2}{x-3}+\dfrac{x-4}{x-5}=\dfrac{10}{3}\]
Now, by using the LCM and adding the fractions on both sides we get
\[\Rightarrow \dfrac{\left( x-2 \right)\left( x-5 \right)+\left( x-3 \right)\left( x-4 \right)}{\left( x-3 \right)\left( x-5 \right)}=\dfrac{10}{3}\]
Now, by multiplying the terms in the numerator and denominator we get
\[\begin{align}
  & \Rightarrow \dfrac{\left( {{x}^{2}}-2x-5x+10 \right)+\left( {{x}^{2}}-3x-4x+12 \right)}{{{x}^{2}}-3x-5x+15}=\dfrac{10}{3} \\
 & \Rightarrow \dfrac{2{{x}^{2}}-14x+22}{{{x}^{2}}-8x+15}=\dfrac{10}{3} \\
\end{align}\]
Now, by cross multiplying the above equation we get
\[\begin{align}
  & \Rightarrow 6{{x}^{2}}-42x+66=10{{x}^{2}}-80x+150 \\
 & \Rightarrow 4{{x}^{2}}-38x+84=0 \\
 & \Rightarrow 2{{x}^{2}}-19x+42=0......equation(i) \\
\end{align}\]
We know that the factorisation method for a quadratic equation of the form
\[a{{x}^{2}}+bx+c=0\]
Then we use the factorisation method that is we take the middle term coefficient \['b'\] as sum of two numbers \[p,q\] such that \[p\times q=a\times c\]
By using the above theorem let us divide ‘-19’ as follows
\[\Rightarrow -19=\left( -12 \right)+\left( -7 \right)\]
Here, we can see that the above equation satisfies the factorisation theorem
By substituting this value in equation (i) we get
\[\Rightarrow \left( 2{{x}^{2}}-12x \right)+\left( -7x+42 \right)=0\]
Now by taking the common terms out we get
\[\begin{align}
  & \Rightarrow 2x\left( x-6 \right)-7\left( x-6 \right)=0 \\
 & \Rightarrow \left( x-6 \right)\left( 2x-7 \right)=0 \\
\end{align}\]
We know that when \[ab=0\] then either \['a'\] or \['b'\] is zero.
By using this result let us take the first term from the above equation that is
\[\begin{align}
  & \Rightarrow x-6=0 \\
 & \Rightarrow x=6 \\
\end{align}\]
Now, by taking the second term we get
\[\begin{align}
  & \Rightarrow 2x-7=0 \\
 & \Rightarrow x=\dfrac{7}{2} \\
\end{align}\]
Therefore the roots of given equation are \[6,\dfrac{7}{2}\]

So, the correct answer is “Option (a)”.

Note: We can check whether the given condition is true or false in other methods.
We are given that the equation as
\[\dfrac{x-2}{x-3}+\dfrac{x-4}{x-5}=\dfrac{10}{3}\]
We are also given that the roots are \[6,\dfrac{7}{2}\]
By substituting \[x=6\] in the LHS of given equation we get
\[\begin{align}
  & \Rightarrow LHS=\dfrac{6-2}{6-3}+\dfrac{6-4}{6-5} \\
 & \Rightarrow LHS=\dfrac{4}{3}+2 \\
 & \Rightarrow LHS=\dfrac{10}{3}=RHS \\
\end{align}\]
Here we can see that LHS is equal to RHS
So, 6 is the root of given equation
Now, by substituting \[x=6\] in the LHS of given equation we get
\[\begin{align}
  & \Rightarrow LHS=\dfrac{\dfrac{7}{2}-2}{\dfrac{7}{2}-3}+\dfrac{\dfrac{7}{2}-4}{\dfrac{7}{2}-5} \\
 & \Rightarrow LHS=\dfrac{3}{1}+\left( \dfrac{-1}{-3} \right) \\
 & \Rightarrow LHS=\dfrac{10}{3}=RHS \\
\end{align}\]
Here we can see that LHS is equal to RHS
So, \[\dfrac{7}{2}\] is the root of given equation
Therefore the roots of given equation are \[6,\dfrac{7}{2}\]
So, option (a) is the correct answer.