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

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

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Answer 1 · 2017-01-15T17:11:41

$a = 5 \pm \sqrt{7}$ $a = 5+-sqrt7$

Explanation:

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

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

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

$a - 5 = \pm \sqrt{7} \Rightarrow a = 5 \pm \sqrt{7}$ $a-5 = +-sqrt7 => a = 5 +- sqrt7$

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

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