Question

FAN AND MEDAL! PLEASE HELP WITH AT LEAST ONE
I have a small quiz in Algebra 2 tomorrow and these are the questions. I don't understand 3-5 Someone please explain these to me step by step

Answer 1

3
square the thing

Answer 2

\[(-2\sqrt6-\sqrt3)(-2\sqrt6-\sqrt3)\] is a first step
then do the usual four multiplications

Answer 3

I know but I checked and I keep getting the wrong answer :/

Answer 4

ok lets do "first" \[(-2\sqrt6)(-2\sqrt6)=?\]

Answer 5

4sqrt 36

Answer 6

hmm what do we call \(\sqrt{36}\) here on earth?

Answer 7

what do you mean?

Answer 8

what is the square root of thirty six?

Answer 9

6

Answer 10

right

Answer 11

you should have not written \(\sqrt{36}\) to begin with, since \[\sqrt6\times \sqrt6=6\] as that is the meaning of \(\sqrt6\), the number you square to get \(6\)

Answer 12

so the "first:" is \(4\times 6=24\)

Answer 13

how bout the "last" \(\sqrt3\times \sqrt3\)?

Answer 14

okay okay that makes sense so far

Answer 15

would that be 3?

Answer 16

of course

Answer 17

square root of 9 is 3

Answer 18

that is what it means, the number you square to get 3

Answer 19

ok so first and last are \(24+3=?\)

Answer 20

okay so I add them both and get 27

Answer 21

yup

Answer 22

now you have to do "outer and inner"

Answer 23

here the square roots do not go away \[2\sqrt6\times \sqrt3=?\]

Answer 24

\[2\sqrt{18}\]

Answer 25

yes, and there is another one, so total is \[4\sqrt{18}\] but not done yet

Answer 26

i don't add the 18s?

Answer 27

?

Answer 28

no, two apples plus two apples is four apples so \[2\sqrt{18}+2\sqrt{18}=4\sqrt{18}\] still not done though

Answer 29

okay

Answer 30

I always mess up on this part

Answer 31

since \(18=9\times 2\) you know \\[\sqrt{18}=\sqrt{9}\sqrt2=3\sqrt2\]

Answer 32

so all together the radical part is \(4\times 3\sqrt2=12\sqrt2\)

Answer 33

omg, you have no Idea how long ive been trying to do this problem

Answer 34

could you help me with four and five?

Answer 35

yeah simplest radical form is a bear

Answer 36

for 4 you have to get the radical out of the denominator
multiply top and bottom by \(\sqrt[3]{3x^2}\) and see what happens

Answer 37

I know about the radical. but how you do get the squared onto the x? Ive been struggling with that rule

Answer 38

you have the cubed root of 9
so if you multiply by 3 you get the cubed root of \(3^3\) which is 3

Answer 39

you have the cubed root of \(x\)
if you multiply by the cubed root of \(x^2\) you get the cubed root of \(x^3\) which is \(x\)

Answer 40

it is not a "rule' you just have to figure it out

Answer 41

second to final answer will be \[\frac{3x\sqrt[3]{3x^2}}{3x}\]

Answer 42

then cancel the \(3x\) top and bottom

Answer 43

last one is the easiest
multply top and bottom by the conjugate of the denominator
the conjugate of \(5-\sqrt{x}\) is \(5+\sqrt{x}\) and this works because \[(5-\sqrt{x})(5+\sqrt{x})=5^2-x\]

Answer 44

Omg thank you so much, you're saving my grade right now

Answer 45

Thanks so much for the help! I really appreciate it