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