Question

Perform the indicated operation. Assume all variables represent non-negative real numbers

$\sqrt{2y^{2} } +y\sqrt{18}$

Answer 1

Answer:

$y\sqrt{2} + 3y\sqrt{2} = 4y\sqrt{2}$

Step-by-step explanation:

To add or subtract radicals, you need to simplify the radicals and then add the radicals

$\sqrt{2y^2} + y\sqrt{18}$

First simplify $\sqrt{2y^2}$

$\sqrt{2y^2}$

$\sqrt{2y^2} = \sqrt{2} \ \sqrt{y^2}$

$\sqrt{2} \ \sqrt{y^2}$

$\sqrt{y^2} = y$

$\sqrt{2} \ y$

OR

$y\sqrt{2}$

Second simplify  $\sqrt{18}$

To simplify  $y\sqrt{18}$ you need to find two radicals that =  $\sqrt{18}$ when you multiply them. One radical needs to be a perfect square and the other needs to be a non perfect square.

$\sqrt{18}$

$\sqrt{9} * \sqrt{2} = \sqrt{18}$

$\sqrt{9} * \sqrt{2}}$

$\sqrt{9} \sqrt{2}}$

$3 \sqrt{2}}$

$3 y\sqrt{2}}$

Now put it all together

$y\sqrt{2} + 3y\sqrt{2} = 4y\sqrt{2}$