Question

Give the exact answer to the problem below. Show algebra which supports your answer.

Answer 1

\[x=\sqrt{6+\sqrt{6+\sqrt{6+...}}}\]\[x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}=6+x\]\[x^2-x-6=0\]

Answer 2

\[x=3\]

Answer 3

WOW O_O

Answer 4

\[x^2-x-6=0\]
\[(x-3)(x+2)=0\]

what does it mean that x=-2 is the other solution ?

Answer 5

thats a good question Unkle

Answer 6

x = 3, x = -2

Answer 7

great work by the way @mukushla

Answer 8

x is the sqrt of a number how can it be negative ?

Answer 9

@UsukiDoll as the square root of any 'real' number cannot be negative, -2 is eliminated
and hence the answer is 3

Answer 10

well yeah if a negative number is in a sq root... it will be imaginary

Answer 11

how did you get a negative inside the sqrt?

Answer 12

O_O negative numbers do show up in square roots. negative numbers are a big no no in square roots so it becomes imaginary.

Answer 13

@UsukiDoll What about the fourth root of -16 ?

Answer 14

dont you guys have a chapter called 'complex numbers'? the real numbers and complex numbers are whole different things XD

Answer 15

dude I'm just trying to solve my own problem .

Answer 16

need to concentrate don't mean to sound rude though

Answer 17

\[\sqrt{-1} = \iota \]

Answer 18

\[\sqrt{6+\sqrt{6+\sqrt{6.......}}}=x\]if we hide one part then the remaining part will be 'x' so we can write it as\[\sqrt{6+x}=x\] now if we square both sides we will get\[x+6=x^2\]\[x^2-x-6=0\]solve this quadratic equation & gewer :)t ur ans

Answer 19

& get ur ans:) *

Answer 20

@UsukiDoll do you believe \[\sqrt{-4} = 2\] ?

Answer 21

no

Answer 22

do you want me to elaborate concept of your confusion?

Answer 23

im going to post a question regarding this one :)

Answer 24

I already know what imaginary numbers look like thanks