sqrt (16x- 16) + sqrt (25x^3- 25x^2)

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sqrt (16x- 16) + sqrt (25x^3- 25x^2)

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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my teacher told us not to like mess with the binomials so im confused
You can factor coefficients out of the first inside expression, so you get: sqrt(16(x-1)) - sqrt(25x^2(x-1)). The 16 in the first term comes to be a 4 on the outside. The 25x^2 inside the second term comes to be a 5x on the outside. 4sqrt(x-1)-5x*sqrt(x-1) And you can simplify by factoring out the sqrt(x-1): sqrt(x-1)*(4-5x).
im so confused

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Do you get that (16x-16)=16(x-1)?
o k now i get it wen u put it simplerr
But 16 is a square so you can bring out it's root
but its connected to a binomial so wich one of them...or both and 25 is also a square connected to a binomial in the laast half of the problem
Well, we rewrite \(\sqrt{16(x-1)}\) as \(\sqrt{16}\sqrt{x-1}\) and simplify.
Can you rewrite the second half now?
yse so the answer is 5x + 4 sqrt x-1?
(4-5x)
and i knock off the sqrt x-1?
no, no, no, just replace the 5x+4 with (4-5x)
o okay
do u mean 4 + 5x

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