Question

Which is equivalent to √75x³-√27x³, if x>0?

Answer 1

`2x√(3x)`

Explanation:

To simplify the expression
`√(75x^3)-√(27x^3)`, we'll want to simplify each radical, and if the two terms become like terms, combine them into a single term.

Some terminology:

For a radical like
`\sqrt[3]{5}`, the "3" (the number written in the little nook) is called the index of the radical, and the "5" (the number/expression) inside of the radical is called the radicand.

Note that for our given expression, the index is not explicitly written for either radical. By default, the index is 2. If we wrote them in explicitly, the expression would look like this:
`\sqrt[2]{75x^3}-\sqrt[2]{27x^3}`

To simplify radicals by hand, factor the radicand (usually completely down to primes), and form groups of matching numbers/constants/variables where the group size is the index of the radical. In this case, the index is "2", so we're looking for pairs of numbers/constants/variables that match after we factor (if it were a cube root, we'd be looking for triplets):

Factoring the first radicand

75 x^3

= 3*25 * x^3

= 3*5*5 * x*x*x

= 3 * (5*5) * x * (x*x)

Factoring the second radicand

27x^3

= 3*9 * x^3

= 3*3*3 * x*x*x

= 3 * (3*3) * x * (x*x)

So, we can rewrite our expression as follows:

`\sqrt[2]{3*(5*5)*x*(x*x)}-\sqrt[2]{3*(3*3)*x*(x*x)}`

Once you've found a group that matches (again, in this case, pairs), it can be factored out of the radical (only bring out one number/letter from each group to represent the group).

From the first radical, we have a pair of 5s, and a pair of "x"s, so it would simplify to
`5x\sqrt[2]{3x}`

Similarly, in the second radical, we have a pair of 3s, and a pair of "x"s, so it would simplify to
`3x\sqrt[2]{3x}`

So, altogether, our expression would simplify to the following at this point...

`5x\sqrt[2]{3x}-3x\sqrt[2]{3x}`

Side note: If everything factored out of the radical, it was a perfect "index" number (in this case, a perfect square number). If this happens, there is no radical left over at the end, and all of the numbers that were factored out, multiply together to make the answer.

Since we're done factoring numbers out of each radical, we don't need the index for each radical written in explicitly to remind us what size group we need, so let's remove them again, and let it be blank to represent the default index of "2".

`5x√(3x)-3x√(3x)`

Notice that both 5, and the negative 3 are multiplied to a short expression, and that that expression,
`x√(3x)`, is exactly the same in both cases. If it's easier to think about, we could call it "c", as it is a common factor of both terms
`c=x√(3x)`

`5(c)-3(c)`

`5c-3c`

`2c`

Substituting back
`c=x√(3x)` ...

`2(x√(3x))`

`2x√(3x)`

So, of the given answer choices (as provided in the comments of the question), the answer would be d.
`2x√(3x)`