Question 335829
There are two properties of radicals which are used when simplifying:<ol><li>{{{root(a, p*q) = root(a, p) * root(a, q)}}}</li><li>{{{root(a, p/q) = root(a, p) / root(a, q)}}}</li></ol>
These properties can be used in both directions and we use both directions when simplifying radicals.<br>
{{{sqrt(r/5) * sqrt(50-r)}}}
Properly simplified square roots do not have square roots in the denominator or fractions within the square root. So we'll start by multiplying the fraction inside the first square root so that it denominator becomes a perfect square:
{{{sqrt((r/5)*(5/5)) * sqrt(50-r)}}}
{{{sqrt(5r/25) * sqrt(50-r)}}}
We can use the second property to separate the first square root:
{{{(sqrt(5r)/sqrt(25)) * sqrt(50-r)}}}
And now we can simplify the denominator:
{{{sqrt(5r)/5 * sqrt(50-r)}}}
And finally we can multiply, usi the first property to multiply the square roots:
{{{(sqrt(5r)* sqrt(50-r))/5}}}
{{{sqrt(5r*(50-r))/5}}}
{{{sqrt(250r-r^2)/5}}}
There are no perfect square factors in this square root (Neither the 25 or the {{{r^2}}} are <b>factors</b> of the expression in the square root!) so this will not simplify any further.<br>
{{{cubed sqrt(-27)}}}
This is called the "cube root of -27". (And to express this using Algebra.com's formula rendering software use: "root(3, -27)" with the 3 braces on each side.)
{{{root(3, -27)}}}
Simplifying radicals is done by finding factors which are powers of the kind of root. With a cube (3rd) root, we are looking for factors which are to the third power (aka perfect cubes). While you may not know many perfect cubes by heart it does not take long to figure them out: 2*2*2 = 8, (-2)*(-2)*(-2) = -8, 3*3*3 = 27, (-3)*(-3)*(-3) = -27, etc. As we can see, {{{-27 = (-3)^3}}} so we can rewrite {{{root(3, -27)}}} as
{{{root(3, (-3)^3)}}}
So we have a cube root of a perfect cube! So
{{{root(3, (-3)^3) = -3}}}<br>
{{{fith sqrt(16x sqared)* fith sqrt (2x 8th y 15th)}}}
These are called "fifth roots". [Algebra.com: root(5, ...)]
{{{root(5, 16x^2)* root(5, 2x^8y^15)}}}
We can use the first property above to multiply these:
{{{root(5, 16x^2*2x^8y^15)}}}
{{{root(5, 32x^10y^15)}}}
Now we look for factors which are powers of 5. It should not take long to find that {{{32 = 2^5}}}. For the variables, we can factor them into powers of 5:
{{{root(5, 2^5*x^5*x^5*y^5*y^5*y^5)}}}
Using the first property, we can separate all these factors:
{{{root(5, 2^5)*root(5, x^5)*root(5, x^5)*root(5, y^5)*root(5, y^5)*root(5, y^5)}}}
All these 5th roots of powers of 5 simplify:
{{{2*x*x*y*y*y}}}
{{{2x^2y^3}}}<br>
{{{sqrt(3x) * sqrt(12x^3)}}}
We can use the first property to multiply these square roots:
{{{sqrt(3x*12x^3)}}}
{{{sqrt(36x^4)}}}
Now we find perfect square factors:
{{{sqrt(6^2*x^2*x^2)}}}
Using the first property again we can separate the prefect square factors:
{{{sqrt(6^2)*sqrt(x^2)*sqrt(x^2)}}}
which simplifies:
{{{6*x*x}}}
{{{6x^2}}}