Which constant must be added and subtracted to complete the square in \(9x^2 + \dfrac{3}{4}x - \sqrt{2} = 0\)?
\(\dfrac{1}{8}\)
\(\dfrac{1}{64}\)
\(\dfrac{1}{4}\)
\(\dfrac{9}{64}\)
Step 1: Look at the quadratic part: \(9x^2 + \dfrac{3}{4}x\).
Factor out \(9\) from these two terms:
\(9x^2 + \dfrac{3}{4}x = 9\Big(x^2 + \dfrac{1}{12}x\Big)\).
Step 2: To complete the square for \(x^2 + \dfrac{1}{12}x\), we use the rule: add \(\Big(\dfrac{coefficient\ of\ x}{2}\Big)^2\).
Here the coefficient of \(x\) is \(\dfrac{1}{12}\).
Step 3: Half of \(\dfrac{1}{12}\) is \(\dfrac{1}{24}\).
Now square it: \(\Big(\dfrac{1}{24}\Big)^2 = \dfrac{1}{576}\).
Step 4: We added \(\dfrac{1}{576}\) inside the bracket, but since the whole bracket is multiplied by 9, the real addition to the expression is:
\(9 \times \dfrac{1}{576} = \dfrac{9}{576} = \dfrac{1}{64}\).
Step 5: Therefore, the constant we add and subtract is \(\dfrac{1}{64}\).
Answer: Option B (\(\dfrac{1}{64}\))