How do you simplify 3y sqrt (4y^3) - sqrt (16y^4) + sqrt( 18y^3)?

May 11, 2015

Before doing any manipulations, let's determine the domain of this expression.
The only "restrictive" operation here is an operation of the square root that is defined in the field of real numbers only for non-negative arguments.
So, the domain is determined by an inequality $y \ge 0$ because only in this case all expressions under the square root operation are defined.

Now we can simplify this expression using a simple transformation that is true for non-negative numbers:
$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ (for any $a \ge 0$, $b \ge 0$)

Using this, notice that
$\sqrt{4 \cdot {y}^{3}} = \sqrt{4} \cdot \sqrt{{y}^{2}} \cdot \sqrt{y} = 2 \cdot y \cdot \sqrt{y}$
$\sqrt{16 {y}^{4}} = \sqrt{16} \cdot \sqrt{{y}^{4}} = 4 \cdot {y}^{2}$
$\sqrt{18 {y}^{3}} = \sqrt{9} \cdot \sqrt{2} \cdot \sqrt{{y}^{2}} \cdot \sqrt{y} = 3 \sqrt{2} \cdot y \cdot \sqrt{y}$

Combining these expression with their corresponding coefficients, produces:
$3 y \sqrt{4 {y}^{3}} - \sqrt{16 {y}^{4}} + \sqrt{18 {y}^{3}} =$
$= 3 \cdot y \cdot 2 \cdot y \cdot \sqrt{y} - 4 \cdot {y}^{2} + 3 \sqrt{2} \cdot y \cdot \sqrt{y} =$
$= 6 {y}^{2} \sqrt{y} - 4 {y}^{2} + 3 \sqrt{2} y \sqrt{y} =$
$= y \sqrt{y} \left(6 y - 4 \sqrt{y} + 3 \sqrt{2}\right)$