help simply 135b^2

Question

would you help me simplify

  ___________        ______

√135b^2c^3d   x  √5b^2d

 

 

 

thanks

Answer 1

is this your equation?

 

(√(135b)^2c)^{^(3d)} x (√(5b))^(2d)

 

We can write squareroots in two ways: as a radical √a or as a power, in which case it would look like this a^1/2

 

might I refer you over to shelovesmath?

she breaks down radicals and exponents very well:

here is a link:

http://www.shelovesmath.com/pre-algebra/exponents-roots-and-scientific-notation/

 

If this is not helpful enough, feel free to email me through wyzant directly and I will respond as quickly as I am able.

 

Answer 2

I am going to attempt to interpret the radicals correctly. However, if you ask a question using radicals in the future it may behoove you to utilize parenthesis instead of attempting to use an over-bar, as formatting does not always work the way you want it to...

 

My guess is that this is intended to be (assuming x is not another variable):

 

√(135b^2c^3d) x √(5b^2d).

 

Now recall that radicals "distribute" over multiplication, and assuming that the "d" variables are not in the powers (and using the superscript command), we get:

 

√(135) x √(b²) x √(c³) x √(d) x √(5) x √(b²) x √(d).

 

Here our expression is very busy, so we will want to clean it up a bit for the sake of reading it. We do this by rearranging terms, and combining any numeric ones we can easily evaluate (but only as far as it is simple to do the evaluation, recalling that we are working on simplification), giving us:

 

√(135 x 5) x √(b²) x √(b²) x √(c³) x √(d) x √(d).

 

Here we intentionally do not evaluate 135 x 5 because it is not simple (and evaluating doesn't lead us to a simplified answer)! Instead, notice that we can factor 135.

 

135

/    \

     5       27

    /     \

  3       9

  /  \

3    3            135=3³ x 5.

 

Notice, too, that √(b²)=|b|, the magnitude of b. And that |b| x |b| = b², giving us:

 

√(3³ x 5 x 5) x (b²) x √(c³) x √(d) x √(d).

 

Here we have in the numeric term √(5 x 5) = √(5) x √(5) = (√(5))² = 5. Likewise √(d) x √(d) = (√(d))² = d, giving us:

 

√(3^3) x 5 x (b^2) x √(c^3) x (d).

 

Furthermore, c^3 = c*c^2, so √(c^3) = √(c x c^2) = c x √(c). The same can be done for √(3^3) by replacing the c's with 3's and writing it all down again, resulting in:

 

3 x √(3) x 5 x (b^2) x c x √(c) x (d).

 

Rearranging again, dropping unnecessary parentheses, and specifically paying attention to group radical and non-radical terms, we have:

 

3 x 5 x b² x c  x d x √(3) x √(c).

 

Rewriting, with radical terms grouped, evaluating were possible (3 x 5 = 15) and dropping multiplication symbols where they are automatically implied, we get:

 

15b²cd√(3c).

 

The parenthesis are are still necessary because I did not use an over-bar.

Answer 3

sqrt(135b^2c^3d)*sqrt(5b^2d)

first just multiply the radicands

sqrt(675b^4c^3d^2)

now separate the number and each variable with  separate radical signs

sqrt(675)sqrt(b^4)sqrt(c^3)sqrt(d^2)

now simplify each radical

sqrt(675)=sqrt(225*3)=15sqrt(3)

sqrt(b^4)=b^2

sqrt(c^3)=sqrt(c^2c^1)=csqrt(c)

sqrt(d^2)=d

15sqrt(3)*b^2*csqrt(c)*d

(15b^2cd)(sqrt[3])(sqrt[c])

15b^2cdsqrt(3c)