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what are the solutions of the equation

Algebra
Tutor: None Selected Time limit: 1 Day

x^2 6x-1=0

Jul 14th, 2014

solve by completing the square.

x² + 6x + 1 = 0
the third term of the square is 9, then replace 1 with (9 - 8):

x² + 6x + 9 -  8 = 0 →

(x² + 6x + 9) -  8 = 0 →
factoring the rising square:

(x + 3)² -  8 = 0 →
finally, factor the left side, viewing it as a difference between
two squares, yielding:

(x + 3)² - (√8)² = 0 →

[(x + 3) + (√8)][(x + 3) - (√8)] = 0 →
(as you know, √8 = √(4)2 = 2√2)

(x + 3 + 2√2)(x + 3 - 2√2) = 0 →
rewrite it in the (x - a)(x - b) form as:

[x - (-3 - 2√2)][x - (-3 + 2√2)] = 0 →
thus the required solutions are:

[x - (-3 - 2√2)] = 0 → x = - 3 - 2√2
and
[x - (-3 + 2√2)] = 0 → x = - 3 + 2√2

Jul 14th, 2014

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