If 6a^2 + 5a + 4 = 3 + 5a^2 + 11a + 5, then what is the smallest possible value of 2a + 1?
+677 6a^2 + 5a + 4 = 3 + 5a^2 + 11a + 5.
\(6a^2 + 5a + 4 = 3 + 5a^2 + 11a + 5\)
Add 3 and 5 together.
\(6a^2 + 5a + 4 = 5a^2 + 11a + 8\)
Subtract 5a^2 + 11a + 8 from both sides.
\(a^2 - 6a - 4 = 0\)
Add 4 from both sides.
\(a^2 - 6a = 4\)
Add 9 to both sides.
\(a^2 - 6a + 9 = 13\)
...
\((a - 3)^2 = 13\)
Take the square root from both sides.
\(a - 3 = \sqrt{13}\)
Add 3 from both sides.
\(a = 3 \pm \sqrt{13}\) >>> \(3 - \sqrt{13} \mbox{ is smaller.}\)
Put \(3 - \sqrt{13}\) in.
\(2a + 1\) >>> \([(3 - \sqrt{13}) \cdot 2] + 1 = \boxed{(6 - 2\sqrt{13}) + 1}\)
