Tell me if I'm right please?

Question

Simplify:
sqrt(252x^5)

I came up with 6x^2*sqrt7x

Answer 1

yes

Answer 2

Note that for positive real numbers,

sqrt(ab) = sqrt(a)sqrt(b),

and this can be done with *any* product.

sqrt (abcde) = sqrt(a)sqrt(b)sqrt(c)sqrt(d)sqrt(e)

For sqrt(252x^5), we obtain the prime factorization.

252 = (2)(126) = (2)(2)(63) = (2)(2)(3)(21) = (2)(2)(3)(3)(7)

Now that we have the factorization, we keep a note of each "pair", and actually multiply them out.

{4}{9}(7)

And we can multiply out the results of the pairs too.

{36} (7)

When decomposing x^5, break off an x so that the power is an even number. x^5 = x^4 * x

sqrt (252x^5) =
sqrt (36 * 7 * x^4 * x)

Keep all non-square terms together; in this case, the 7 and x.

sqrt (36 * x^4 * 7x)

Now, apply that square root property we mentioned earlier.

sqrt(36) sqrt(x^4) sqrt(7x)

Note that square root of 36 is 6, and the square root of x^4 is x^2. Our answer is then

6 (x^2) sqrt(7x)

So you are indeed correct. Pardon my verboseness.

Answer 3

You are: expressed with factors that are perfect squares, you get:
sqrt(36*7*x^4*x)
Then taking the square root of the perfect squares:
6x^2(sqrt(7x)

Answer 4

Yes but with parentheses for order of ops 6x^2*sqrt(7x)

Answer 5

You're right on. Good job!