Question

Solve the given equation
$\dfrac{1}{2}\left( 1-\dfrac{2}{m} \right)\left( 2-m \right)=0$

Answer 1

Hint: First use distributive law to expand the equation in terms of m. Now take the least common multiple to get a quadratic in m. By factoring the quadratic, find the value of the m. The value of m is the required result in the question.

Complete step-by-step answer:
Given in the question is written in form of
$\dfrac{1}{2}\left( 1-\dfrac{2}{m} \right)\left( 2-m \right)=0$
By assuming the term (2 – m) as k, we get equation as
$\dfrac{1}{2}\left( 1-\dfrac{2}{m} \right)k=0$
By using distributive law, we can write it the equation as
$\dfrac{k}{2}-\left( \dfrac{k}{2} \right)\left( \dfrac{2}{m} \right)=0$
By simplifying the above equation, we can write it as
$\dfrac{k}{2}-\dfrac{k}{m}=0$
By substituting the value of k back into equation, we get
$\left( \dfrac{2-m}{2} \right)-\dfrac{1}{m}\left( 2-m \right)=0$
By using distributing law on second term of equation
$1-\dfrac{m}{2}-\dfrac{2}{m}+1=0$
By simplifying the above equation, we can write it as
$\dfrac{-m}{2}-\dfrac{2}{m}+2=0$
By taking least common multiple, we can write it as
$\dfrac{\left( -m \right)\left( m \right)+\left( -2 \right)\left( 2 \right)+\left( 2 \right)\left( 2m \right)}{2m}=0$

By doing cross multiplication, we can write the equation as
$-{{m}^{2}}-4+4m=0$
By taking -1 common from whole equation, we get it as
$-1\left( {{m}^{2}}+4-4m \right)=0$
By dividing with -1 on both sides, we get it in form of
${{m}^{2}}-4m+4=0$
By writing -4m as - 2m - 2m we can write equation as
${{m}^{2}}-2m-2m+4=0$
By taking m common from first two terms, we get it as
\[m(m-2)-2m+4=0\]
By taking -2 common from last two terms, we get it as
\[m(m-2)-2(m-2)=0\]
By taking \[(m-2)\] common from whole equation we get it as
\[(m-2)(m-2)=0\]
By equating each of it to 0, we can write it as
\[m=2,2\]
So, the solution m satisfying the above equation can be written as 2.

Note: Be careful while using distributive law be careful with “-“ signs in distributive law the only way to make mistakes is taking “-“ as “+” or vice versa. Alternatively you can use the quadrant formula to get the value of m. Hence discriminant is zero, so however you get the same roots. Another method is you can use the algebraic identity of ${{\left( a-b \right)}^{2}}$ at the quadratic step or you can directly send m to RHS in the first step. Then also you get the same roots as solutions.