How do you solve $x^2 - 18x + 74 = 0$ by completing the square?

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Answer 1 · 2016-06-13T06:21:25

$x=9+-sqrt(7)$

Complete the square and use the difference of squares identity:

$a^2-b^2=(a-b)(a+b)$

with $a=(x-9)$ and $b=sqrt(7)$ as follows:

$0 = x^2-18x+74$

$= x^2-18x+81-7$

$= (x-9)^2-(sqrt(7))^2$

$= ((x-9)-sqrt(7))((x-9)+sqrt(7))$

$= (x-9-sqrt(7))(x-9+sqrt(7))$

Hence the zeros are:

$x = 9 +-sqrt(7)$

How do you solve x^2 - 18x + 74 = 0x^2 - 18x + 74 = 0 by completing the square?