# Will give 20 points When you have a square root of a number, you essentially have to find a number that can be multiplied by itself to be equivalent to the original number.Take, for example, $$\sqrt[]{4}$$. 2 times 2 is 4, meaning that the square root of 4 is 2.Back to the question.2$$\sqrt[]{18}$$ - 5$$\sqrt[]{32}$$First, let's find factors of the numbers that have to be squared.2$$\sqrt[]{9 * 2}$$ - 5$$\sqrt[]{16 * 2}$$Neat! It seems like we already have two squares available!2$$\sqrt[]{(3 * 3) * 2}$$ - 5$$\sqrt[]{(4 * 4) * 2}$$Square 3 and 4:2*3$$\sqrt[]{2}$$ - 5*4$$\sqrt[]{2}$$6$$\sqrt[]{2}$$ - 20$$\sqrt[]{2}$$-14$$\sqrt[]{2}$$, Option C.Now, onto the second question.7$$\sqrt[]{24}$$ + $$\sqrt[]{90}$$ - 8$$\sqrt[]{54}$$Like the first question, let's find some factors7$$\sqrt[]{4 * 6}$$ + $$\sqrt[]{9 * 10}$$ - 8$$\sqrt[]{9 * 6}$$Aha! Looks like we found a couple more squares!7$$\sqrt[]{(2*2) * 6}$$ + $$\sqrt[]{(3*3) * 10}$$ - 8$$\sqrt[]{(3*3) * 6}$$Square all of them:7*2$$\sqrt[]{6}$$ + 3$$\sqrt[]{10}$$ - 8*3$$\sqrt[]{6}$$14$$\sqrt[]{6}$$ + 3$$\sqrt[]{10}$$ - 24$$\sqrt[]{6}$$Note that you can only add numbers with similar leftover roots, so let's do just that:(14$$\sqrt[]{6}$$ - 24$$\sqrt[]{6}$$) + 3$$\sqrt[]{10}$$-10$$\sqrt[]{6}$$ + 3$$\sqrt[]{10}$$, Option D.If you have any questions, don't be afraid to ask! Good luck :))-T.B.