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IgorLugansk [536]
3 years ago
13

Solve for (x-4)^2=13

Mathematics
2 answers:
ryzh [129]3 years ago
8 0
(x-4)^2=13 
First take the square root of both sides. 
x-4= \sqrt{13}, - \sqrt{13}. There are two solutions because a square root comes with a positive and negative solution.
Now move the 4 over by adding 4 on both sides. 
x= \sqrt{13}+4, - \sqrt{13} +4

Norma-Jean [14]3 years ago
6 0
                         (x - 4)² = 13
                 (x - 4)(x - 4) = 13
          x(x - 4) - 4(x - 4) = 13
x(x) - x(4) - 4(x) + 4(4) = 13
          x² - 4x - 4x + 16 = 13
                 x² - 8x + 16 = 13
                             - 13  - 13
                    x² - 8x + 3 = 0
                    x = -(-8) ± √((-8)² - 4(1)(3))
                                        2(1)
                    x = 8 ± √(64 - 12)
                                    2
                    x = 8 ± √(52)
                                2
                    x = 8 ± 2√(13)
                                 2
                    x = 4 ± √(13)
                    x = 4 + √(13)    or    x = 4 - √(13)
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