Question 374416
{{{sqrt(175y^6 )}}}
= {{{sqrt(175*y^2 )}}}
Where did the 6 go??<br>
To simplify a square root (where there are no fractions), like this one, you look for perfect square factors of the radicand. (The expression inside a radical is called the radicand.) Your radicand has a couple of perfect square factors:
{{{sqrt(25*7*(y^3)^2)}}}
Next I like to use the Commutative Property to rearrange the factors so the perfect squares are in front:
{{{sqrt(25*(y^3)^2*7)}}}
Next we use a property of radicals, {{{root(a, p*q) = root(a, p)*root(a, q)}}}, to separate all the perfect square factors into their own square roots:
{{{sqrt(25)*sqrt((y^3)^2)*sqrt(7)}}}
The square roots of the perfect squares can be simplified:
{{{5*y^3*sqrt(7)}}}
One more detail...
Square roots, including {{{sqrt(175y^6 )}}}, are supposed to be positive (or zero). Our simplified expression should also be positive (or zero), too. However,
{{{5*y^3*sqrt(7)}}}
could be negative if y is negative. Since we do not know that y cannot be negative, then we must use absolute value to ensure a positive (or zero) expression:
{{{5*abs(y^3)*sqrt(7)}}}
Since {{{y^2}}} cannot be negative this can also be written as
{{{5*y^2*abs(y)*sqrt(7)}}}
Either of these last two is your simplified expression.