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# help algebra

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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.

$$6a^2 + 5a + 4 = 3 + 5a^2 + 11a + 5$$

$$6a^2 + 5a + 4 = 5a^2 + 11a + 8$$

Subtract 5a^2 + 11a + 8 from both sides.

$$a^2 - 6a - 4 = 0$$

$$a^2 - 6a = 4$$

$$a^2 - 6a + 9 = 13$$

...

$$(a - 3)^2 = 13$$

Take the square root from both sides.

$$a - 3 = \sqrt{13}$$

$$a = 3 \pm \sqrt{13}$$  >>> $$3 - \sqrt{13} \mbox{ is smaller.}$$

Put $$3 - \sqrt{13}$$ in.

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

Dec 31, 2021