Question

http://www4b.wolframalpha.com/Calculate/MSP/MSP5901a36260gc491c9de00002gfgi1a471hifeg0?MSPStoreType=image/gif&s=63&w=194&h=28

Plaintext: sqrt(3+sqrt(8))^x+sqrt(3-sqrt(8))^x = 6

Answer 1

\[(3+\sqrt{8})(3-\sqrt{8})=1\]

Answer 2

How did you get that?

Answer 3

(a-b)(a+b)=a^2-b^2 right??

Answer 4

yes but the a + b and a - b you're referring to are both individually to the x power

Answer 5

is that\[(\sqrt{3+\sqrt{8}})^x+(\sqrt{3-\sqrt{8}})^x=6\]

Answer 6

yes

Answer 7

and are u agree with\[\sqrt{3+\sqrt{8}} \times \sqrt{3-\sqrt{8}}=1\]?

Answer 8

How did you just get rid of the ^x's?

Answer 9

it has nothing to do with equation...its an other thing we would use for solving equation

Answer 10

how do you know that then?

Answer 11

\[\sqrt{3+\sqrt{8}} \times \sqrt{3-\sqrt{8}}=\sqrt{(3+\sqrt{8})(3-\sqrt{8})}=\sqrt{3^2-(\sqrt{8})^2}=\sqrt{9-8}=1\]

Answer 12

oh okay i get that

Answer 13

sorry i gotta go...\[\sqrt{3-\sqrt{8}}=\frac{1}{\sqrt{3+\sqrt{8}}}\]put this in the equation\[(\sqrt{3+\sqrt{8}})^x+(\frac{1}{\sqrt{3+\sqrt{8}}})^x=6\]\[(\sqrt{3+\sqrt{8}})^x+\frac{1}{(\sqrt{3+\sqrt{8}})^x}=6\]now let\[y=(\sqrt{3+\sqrt{8}})^x\]\[y+\frac{1}{y}=6\]u have a quadratic again

Answer 14

how did you get the first step?

Answer 15

he used the idea that when multiplied together you get 1
solve for one in terms of the other

Answer 16

thank you (both)!

Answer 17

i've gotten up to http://www4b.wolframalpha.com/Calculate/MSP/MSP34481a36271371i95feg000017g5b3h992504fh8?MSPStoreType=image/gif&s=33&w=239&h=31
i solved the 3+ 2(sqrt)2 part but need help on the 3 - (sqrt)2 part

Answer 18

3 - 2(sqrt)2=(3+ 2(sqrt)2)^(-1)

right?

Answer 19

i thought it was:

3 - 2(sqrt)2=(3+ 2(sqrt)2)^(x)

Answer 20

i mean

3 - 2(sqrt)2=(3+ 2(sqrt)2)^(-1)

"also"

Answer 21

ref --> my 6th reply

Answer 22

oh right

Answer 23

my first thought is get rid of the square root, and use x/2 as the exponent
also write sqrt(8) as 2sqrt(2)

Answer 24

then take ln of both sides. one solution looks like an integer

Answer 25

@mukushla for the 3 - 2(sqrt)2 part I got x = -1, but i got x = 2 for the 3 + 2(sqrt)2 part so i thought it would be -2...?

Answer 26

its -2 check again...and as @phi mentioned get rid of the square root, and use x/2 as the exponent

Answer 27

oh i got it you can square both sides and since the bases are equal set the exponents equal so x = -2 thank you both so much

Answer 28

np :)

Answer 29

@phi i got it thank you