Question

please help

Answer 1

Help.?

Answer 2

multiply top and bottom times the conjugate of the denominator.

Answer 3

How Do You Attach A Picture To The Post.?

Answer 4

click "attach file" below

Answer 5

saarupp, do you know what a conjugate for the denominator i this case is going to be?

Answer 6

Conjugate for any \(\large\color{slate}{ a-b }\), is \(\large\color{royalblue}{ a+b }\).

Answer 7

hello, are you there? :)

Answer 8

I'm so lost

Answer 9

Me Too.?

Answer 10

the rule is: \(\large\color{teal}{ (a-b)(a+b)=a^2-b^2 }\)

that means that any square roots, in the expression, \(\large\color{royalblue}{ (3-\sqrt{2})(3+\sqrt{2}) }\) will be raised to the second power, and any roots disappear.

Answer 11

based on the afore mentioned rule, can you expand the: \(\large\color{royalblue}{ (3-\sqrt{2})(3+\sqrt{2}) }\) ?

Answer 12

9?

Answer 13

3 times 3 is 9. so that would be the first term. it is the "a^2" in this case.

Answer 14

and \(\large\color{forestgreen}{ \sqrt{2} \times \sqrt{2} }\) is equivalent to what?

Answer 15

sqrt 4?

Answer 16

yes, and that is same as?

Answer 17

2

Answer 18

yup

Answer 19

So,
\(\large\color{royalblue}{ (3-\sqrt{2})(3+\sqrt{2}) =9-2 }\)

Answer 20

so the bottom is 7. (because 9-2, is equal to 7, we all know that)

Answer 21

what happens when you multiply: \(\large\color{royalblue}{ 9\times(3+\sqrt{2}) }\) ?

Answer 22

27 +sqrt18?

Answer 23

you got the 27 corrrectly.

Answer 24

\(\large\color{royalblue}{ 9\times \sqrt{2} }\) will be just \(\large\color{royalblue}{ 9\sqrt{2} }\). you can't reduce it

Answer 25

so it's 27+9 sqrt2/7?

Answer 26

yes

Answer 27

you are correct, the answer is B

Answer 28

Thank you!! so much

Answer 29

it is not a problem!