How do you solve $a^2 - 10a +18 = 0$ by completing the square?

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Answer 1 · 2017-01-15T17:11:41

$a = 5+-sqrt7$

Since $(a-b)^2 = a^2 - 2ab + b^2$ , we need the middle term to be

$-2ab$ , so we use $5$ as $b$ .

$a^2 - 10a + 25 = (a - 5)^2$ , so $(a-5)^2 - 7 = 0 => (a-5)^2 = 7$

$a-5 = +-sqrt7 => a = 5 +- sqrt7$

How do you solve a^2 - 10a +18 = 0a^2 - 10a +18 = 0 by completing the square?