Complete the square x^2+6x=7

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Complete the square x^2+6x=7

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[x^2+6x-7=0\] \[(x+7)(x-1)=0\]
x^2+6x =7 remember x^2 = x^2+2ax+a^2 x^2 is x^2. 6x is 2ax, we have to find a we can find a by dividing 6x by 2, which is 3. So a =3 but remember a^2, so 3^2 is 9 (3 times 3 equals 9) now you can plug that in. x^2+6x+9=7 but what we do to one side we must do to the other x^2+6x+9=7+9 right off the bat we can add 7 and 9 x^2+6x+9=16 ok now, lets factor it out by using difference of two perfect squares (you can ignore 16 for now) (x+3)(x+3) now you can bring back the 16 (x+3)^2=16 btw (x+3)^2 is same as (x+3)(x+3) now, we have to find the square root of (x+3)^2 and 16 the square root of (x+3)^2 is just (x+3) and the square root of 16 is just 4 so (x+3) = (+-) 4 remember the plus or minus sign with the square roots now , subtract 3 from both sides x= -3 (+-) 4 now just do basic arithmetic x = -3 + 4 x= 1 is one solution x = -3 - 4 x= -7 is another solution
http://www.wolframalpha.com/input/?i=x%5E2%2B6x%3D7

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Other answers:

\[x^2+6x-7=0\]\[(x+3)^2-3^2-7=0\]\[(x+3)^2-16=0\]\[(x+3)^2=16\]\[x+3=\pm 4\]\[x=1 , -7\]

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