How do you complete the square to solve $x^{2} - 2 x - 35 = 0$ $x^2 - 2x - 35 = 0$ ?

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Answer 1 · 2015-05-25T15:46:27

$0 = x^{2} - 2 x - 35$ $0 = x^2-2x-35$

$= x^{2} - 2 x + 1 - 1 - 35$ $= x^2-2x+1-1-35$

$= {(x - 1)}^{2} - 36$ $=(x-1)^2-36$

$= {(x - 1)}^{2} - 6^{2}$ $= (x-1)^2-6^2$

$= ((x - 1) - 6) ((x - 1) + 6)$ $= ((x-1)-6)((x-1)+6)$

$= (x - 7) (x + 5)$ $= (x-7)(x+5)$

So $x = 7$ $x=7$ or $x = - 5$ $x=-5$

I used the difference of squares inequality along the way:

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

How do you complete the square to solve x^2 - 2x - 35 = 0x2−2x−35=0x^2 - 2x - 35 = 0?