Question

I don't understand question 3c in the following post

Answer 1

What is the equation in question?

Answer 2

the one in part b

Answer 3

so, you got till t+1/t=4
right ?
now multiply both sides by 't' , what u get ?

Answer 4

t^2 + 1 = 4t

Answer 5

so, can u solve the quadratic equation
t^2-4t+1=0 ??

Answer 6

(t-2)^2 - 3

Answer 7

(t-2)^2 = 3
take square root on both sides.

Answer 8

actually u get
\(t - 2 =\pm \sqrt{(3)}\)

Answer 9

yes my mistake

Answer 10

which completes part c

Answer 11

have douts on part d also ?

Answer 12

*doubts

Answer 13

i thought i had to sub in t from the previous answer

Answer 14

and i didnt understand how to solve for x

Answer 15

yes.
\(t =2\pm \sqrt{(3)} \\ (\sqrt{2+\sqrt3})^x=2\pm \sqrt{(3)}\)

Answer 16

so x = 2

Answer 17

\( (\sqrt{2+\sqrt3})^x=2\pm \sqrt{(3)} \\ (\sqrt{2+\sqrt3})^x=2+ \sqrt{(3)} \\ and \\ (\sqrt{2+\sqrt3})^x=2- \sqrt{(3)}\)

Answer 18

for
\( (\sqrt{2+\sqrt3})^x=2+ \sqrt{(3)}\)
you get x=2, correct

Answer 19

what about \( (\sqrt{2+\sqrt3})^x=2- \sqrt{(3)}\)
?

Answer 20

2-sqrt3 = 1/(2+sqrt3) is that right

Answer 21

yes, use that.

Answer 22

2-sqrt3 = 1/(2+sqrt3) = (2+sqrt3)^(-1)

Answer 23

x = -3

Answer 24

-3 ?? O.o

Answer 25

how ?

Answer 26

wait its -2

Answer 27

\(\checkmark\)
good :)

Answer 28

thanks for your help, you're a lifesaver ;D

Answer 29

welcome ^_^