√(x+16)=x+4

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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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\[\sqrt{x+16}=x+4\]
square both sides x+16=(x+4)^2 (x+4)^2 is the same as (x+4)(x+4)= x^2+8x+16 x+16=x^2+8x+16 subtract both sides by 16 x=x^2+8x subtract both sides by x 0=x^2+8x-x 8x-x=7x 0=x^2+7x Factor out an x 0=x(x+7) Now you have to find at what value of x the equation is equal to 0. The obvious one is x=0 because: 0*(0+7) is 0 The other one is x=(-7) -7*(-7+7)=-7*(0)=0 So x= -7, 0
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x=0 is correct. Radical 16 = 4. at x=-7, 3 radical 9 is not equal to -3
"3 radical 9" should have been posted as "radical 9"
wait WHAT?
There is no negative sign, explicit or inferred as far as I can see, in front of the radical sign in your problem statement. If you replace x with -7 in your problem equation and do the math, you will get on the left hand side, \[\sqrt{-7+16}=3 \] On the right and side you will get -7+4=-3. 3 is unequal to -3. In other words -7 is not a solution to the problem as presented.

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