Answer: The correct option is (c) [tex]\dfrac{49}{100}.[/tex]
Step-by-step explanation: We are given to solve the following quadratic equation by the method of completing the square:
[tex]5x^2-7x+2=0~~~~~~~~~~~~~~~~~~~(i)[/tex]
Also, we are to find the constant added on both sides to form the perfect square trinomial.
We have from equation (i) that
[tex]5x^2-7x+2=0\\\\\Rightarrow x^2-\dfrac{7}{5}x+\dfrac{2}{5}=0\\\\\\\Rightarrow x^2-2\times x\times \dfrac{7}{10}+\left(\dfrac{7}{10}\right)^2+\dfrac{2}{5}=\left(\dfrac{7}{10}\right)^2\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49}{100}-\dfrac{2}{5}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49-40}{100}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{9}{100}\\\\\\\Rightarrow x-\dfrac{7}{10}=\pm\dfrac{3}{10}\\\\\\\Rightarrow x=\pm\dfrac{3}{10}+\dfrac{7}{10}.[/tex]
So,
[tex]x=\dfrac{3}{10}+\dfrac{7}{10},~~~~~~~~x=-\dfrac{3}{10}+\dfrac{7}{10}\\\\\\\Rightarrow x=\dfrac{10}{10},~~~~~~~~\Rightarrow x=\dfrac{-3+7}{10}\\\\\\\Rightarrow x=1,~-\dfrac{2}{5}.[/tex]
Thus, the required solution is [tex]x=1,~-\dfrac{2}{5}.[/tex] and the value of the constant added is [tex]\dfrac{49}{100}.[/tex]
Option (c) is correct.
