Question

How do you solve for \[x\] in \[\left( x-5 \right)\left( x-6 \right)=\dfrac{25}{{{24}^{2}}}\]?

Answer 1

Hint: In order to solve the given question \[\left( x-5 \right)\left( x-6 \right)=\dfrac{25}{{{24}^{2}}}\] and find the value of \[x\], First assume the terms in bracket be equal to a variable. Here a “variable” is a symbol which functions as a placeholder for varying expressions or quantities. Then substitute this variable to the given equation. After this simplify the quadratic equation with the help of splitting the middle term and expanding the brackets. You’ll get the value of the assumed variable. Now, at last substitute the value to the bracket term to find the value of \[x\]

Complete step by step answer:
According to the question given equation is as follows:
\[\left( x-5 \right)\left( x-6 \right)=\dfrac{25}{{{24}^{2}}}...\left( 1 \right)\]
In order to find the value for \[x\], assume the terms in bracket be equal to a variable that is:
Let \[\left( x-6 \right)=a\];
 then \[\left( x-5 \right)=a+1\]
Now substitute these above values in equation \[\left( 1 \right)\] we get:
\[\Rightarrow \left( a+1 \right)a=\dfrac{24+1}{{{24}^{2}}}\]
Now after taking \[\dfrac{1}{24}\] common from the right-hand side of the expression we get:
\[\Rightarrow \left( a+1 \right)a=\dfrac{1}{24}\left( 1+\dfrac{1}{24} \right)\]
Take all the terms to the left-hand side of the equation, we’ll get the equation in quadratic form that is :
\[\Rightarrow {{a}^{2}}+a-\dfrac{1}{24}\left( 1+\dfrac{1}{24} \right)=0\]
Apply the concept of splitting the middle term, to solve the quadratic equation:
\[\Rightarrow {{a}^{2}}+\left( 1+\dfrac{1}{24} \right)a-\dfrac{a}{24}-\dfrac{1}{24}\left( 1+\dfrac{1}{24} \right)=0\]
Observe the above equation and take the terms in common and rest in brackets we get:
\[\Rightarrow a\left( a+1+\dfrac{1}{24} \right)-\dfrac{1}{24}\left( a+1+\dfrac{1}{24} \right)=0\]
Repeat the above step again that is take the terms in common and rest in brackets, we get:
\[\Rightarrow \left( a+1+\dfrac{1}{24} \right)\left( a-\dfrac{1}{24} \right)=0\]
After simplifying it further we get:
\[\Rightarrow \left( a+\dfrac{25}{24} \right)\left( a-\dfrac{1}{24} \right)=0\]
So, values are as follows:
 \[~a=-\dfrac{25}{24}\] and \[a=\dfrac{1}{24}\]
As you can see there are two values of \[a\] which implies there will be two cases where we’ll find x for each value of \[a\].
Case 1: When \[a=-\dfrac{25}{24}\]
\[\Rightarrow x=a+6=-\dfrac{25}{24}+6\]
\[\Rightarrow x=4\dfrac{23}{24}\]
Case 2: When \[a=\dfrac{1}{24}\]
\[\Rightarrow x=a+6=\dfrac{1}{24}+6\]
\[\Rightarrow x=6\dfrac{1}{24}\]

Therefore, values for \[x\] in the given equation \[\left( x-5 \right)\left( x-6 \right)=\dfrac{25}{{{24}^{2}}}\] are \[4\dfrac{23}{24}\]and \[6\dfrac{1}{24}\].

Note: Students generally make mistakes while applying the concept of splitting the middle term to solve the quadratic term, they make mistakes in identifying the common terms and take wrong terms in common which further leads to the wrong answer. Key point is to remember this concept of splitting the middle term that is in quadratic factorization, Splitting of Middle Term is variable term is the sum of two factors and product equal to last term.