Discussion
If \(a + b = \frac{17}{6}\) and ab = 2, then what is the value of (a − b), where a > b ?
Answer: A
The value of (a − b) is \(\frac{1}{6}\).
Use the identity \((a - b)^2 = (a + b)^2 - 4ab\).
\[ \begin{aligned} (a - b)^2 &= \left(\frac{17}{6}\right)^2 - 4(2) \\ &= \frac{289}{36} - \frac{288}{36} = \frac{1}{36} \end{aligned} \]
Since a > b, a − b is positive, so it equals \(\frac{1}{6}\). Check: a = 3/2 and b = 4/3 add to 17/6, multiply to 2 and differ by 1/6. Stopping at \((a - b)^2 = \frac{1}{36}\) without taking the square root is the trap; 1/12, 1/3 and 2/3 do not square to 1/36.
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