\(\frac{49}{24}\)
\(\frac{49}{12}\)
\(\frac{49}{13}\)
\(\frac{49}{3}\)
\(\frac{49}{36}\)
Correct answer is B
3x2 - 7x + R. Computing the square, we have
x2 - \(\frac{7}{3}\) = -\(\frac{R}{3}\)
(\(\frac{x}{1} - \frac{7}{6}\))2 = -\(\frac{R}{3}\) + \(\frac{49}{36}\)
\(\frac{-R}{3}\) + \(\frac{49}{36}\) = 0
R = \(\frac{49}{36}\) x \(\frac{3}{1}\)
= \(\frac{49}{12}\)
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