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If 6a^2 + 5a + 4 = 3 + 5a^2 + 11a + 5, then what is the smallest possible value of 2a + 1?

 Dec 31, 2021
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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.

 

\(2a + 1\) >>> \([(3 - \sqrt{13}) \cdot 2] + 1 = \boxed{(6 - 2\sqrt{13}) + 1}\)

 Dec 31, 2021

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