Radical and Rational Exponent

find roots
square roots of 12a^3/25=6a^3

-3-square roots 18/-6=-1

check this for me it find the roots of the problem.

To find the roots of the equation, we first need to simplify the expression on the left-hand side:

The given equation is:

(sqrt(12a^3)/25) = 6a^3

To simplify, let's first simplify the square root part. The square root of 12a^3 can be written as (12a^3)^(1/2):

(12a^3)^(1/2)/25 = 6a^3

Next, we can simplify the expression inside the parentheses. Using the property of exponents that (a^m)^n = a^(m*n), we have:

12^(1/2) * (a^3)^(1/2) / 25 = 6a^3

Taking the square root of 12 gives us:

2 * a^(3/2) / 25 = 6a^3

Now let's solve for a. To get rid of the fraction, we can multiply both sides by 25:

2 * a^(3/2) = 6a^3 * 25

Simplifying further, we have:

2 * a^(3/2) = 150a^3

Next, let's bring all the terms with a to one side and simplify:

2 * a^(3/2) - 150a^3 = 0

To solve this equation, we can set each term equal to zero and solve for a. Thus, we have two equations:

2 * a^(3/2) = 0
-150a^3 = 0

The first equation, 2 * a^(3/2) = 0, is solved by setting a^(3/2) = 0. To solve this equation, we can raise both sides of the equation to the power of (2/3):

(a^(3/2))^(2/3) = 0^(2/3)

This simplifies to:

a = 0

Therefore, the equation 2 * a^(3/2) = 0 has one root, a = 0.

The second equation, -150a^3 = 0, is solved by setting -150a^3 = 0. To solve this equation, we divide both sides of the equation by -150:

a^3 = 0

Taking the cube root of both sides, we get:

a = 0

Therefore, the equation -150a^3 = 0 also has one root, a = 0.

To summarize, the equation sqrt(12a^3)/25 = 6a^3 has two roots: a = 0.