Question

How do you solve for $x$? $\left( x-2 \right)\left( x-3 \right)=\dfrac{34}{{{33}^{2}}}$

Answer 1

Hint: For solving the equation $\left( x-2 \right)\left( x-3 \right)=\dfrac{34}{{{33}^{2}}}$, we need to write the RHS as the product of $\dfrac{34}{33}$ and $\dfrac{1}{33}$. The difference between the factors on the LHS is the same as the difference between the numbers $\dfrac{34}{33}$ and $\dfrac{1}{33}$. So by comparing the LHS and RHS, we can equate the greater factor with the greater number to get a solution of the equation. For the other solution, we have to write the RHS of the given equation as the product of $-\dfrac{34}{33}$ and follow the same procedure.

Complete step by step answer:
The equation given in the above question is written as
$\left( x-2 \right)\left( x-3 \right)=\dfrac{34}{{{33}^{2}}}$
Now, we can see that the above equation consists of two linear factors in the variable $x$. This means that the above equation is a quadratic equation in the variable $x$. So there will be at most two solutions of the given equation.
Now, the RHS of the above equation can also be written as
$\Rightarrow \left( x-2 \right)\left( x-3 \right)=\dfrac{34}{33}\times \dfrac{1}{33}$
The LHS of the above equation suggests that the multiplication of the two factors $\left( x-2 \right)$ and $\left( x-3 \right)$ is equal to the multiplication of the two numbers \[\dfrac{34}{33}\] and \[\dfrac{1}{33}\]. So if we can write the two numbers in terms of the two factors, we can solve the above equation by comparison. Let us find the difference between the two factors as
\[\begin{align}
  & \Rightarrow d=\left( x-2 \right)-\left( x-3 \right) \\
 & \Rightarrow d=x-2-x+3 \\
 & \Rightarrow d=1 \\
\end{align}\]
So the factor \[\left( x-2 \right)\] is greater than the factor \[\left( x-3 \right)\] by one.
So difference between the factors is equal to one. Now, let us calculate the difference between the numbers \[\dfrac{34}{33}\] and \[\dfrac{1}{33}\] as
\[\begin{align}
  & \Rightarrow D=\dfrac{34}{33}-\dfrac{1}{33} \\
 & \Rightarrow D=\dfrac{33}{33} \\
 & \Rightarrow D=1 \\
\end{align}\]
So the number \[\dfrac{34}{33}\] is greater than the number \[\dfrac{1}{33}\] by one.
Therefore, by comparing the LHS and RHS of the equation $\left( x-2 \right)\left( x-3 \right)=\dfrac{34}{33}\times \dfrac{1}{33}$, we can equate the greater factor to the greater number to get
\[\begin{align}
  & \Rightarrow x-2=\dfrac{34}{33} \\
 & \Rightarrow x=2+\dfrac{34}{33} \\
 & \Rightarrow x=\dfrac{100}{33} \\
\end{align}\]
So one solution of the given equation is $x=\dfrac{100}{33}$.
For the second solution, we write the RHS of the given equation as
$\Rightarrow \left( x-2 \right)\left( x-3 \right)=\left( \dfrac{-34}{33} \right)\times \left( \dfrac{-1}{33} \right)$
In this case, the greater number is $\dfrac{-1}{33}$. So we equate it to the greater factor $\left( x-2 \right)$ to get
\[\begin{align}
  & \Rightarrow x-2=\dfrac{-1}{33} \\
 & \Rightarrow x=2-\dfrac{1}{33} \\
 & \Rightarrow x=\dfrac{65}{33} \\
\end{align}\]
So the second solution of the given equation is \[x=\dfrac{65}{33}\].

Hence, the solutions of the equation $\left( x-2 \right)\left( x-3 \right)=\dfrac{34}{{{33}^{2}}}$ are $x=\dfrac{100}{33}$ and \[x=\dfrac{65}{33}\].

Note: We can also write the given quadratic equation in the standard form of $a{{x}^{2}}+bx+c=0$ and apply the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the above question. But applying the quadratic formula for the equation $\left( x-2 \right)\left( x-3 \right)=\dfrac{34}{{{33}^{2}}}$ is quite cumbersome. So for these types of questions, the comparison method must only be preferred.