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