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Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^2+6x7=0\] \[(x+7)(x1)=0\]

neverforgetvivistee
 3 years ago
Best ResponseYou've already chosen the best response.1x^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

Sarah.L
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^2+6x7=0\]\[(x+3)^23^27=0\]\[(x+3)^216=0\]\[(x+3)^2=16\]\[x+3=\pm 4\]\[x=1 , 7\]
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