Question

If \[x=\sqrt{3+\sqrt{3+\sqrt{3+....}}}\], then,

Answer 1

\[x^2 = 3 + x\]

Answer 2

now solve for x

Answer 3

\[Ans. 2<x <3\]

Answer 4

I tried this but i don't get it:(

Answer 5

do you get \[x^2 = 3+x\]

Answer 6

Ya!

Answer 7

\[x^2 - x - 3 = 0\]
use Q.F>
\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)}\]
\[x = \frac{1 \pm \sqrt{1 + 12}}{2}\]
\[x = \frac{1 \pm \sqrt{13}}{2}\]

Answer 8

then get the positive answer

Answer 9

you will get something around \[x = 2.3\]
got it?

Answer 10

@jiteshmeghwal9

Answer 11

No @lgbasallote

Answer 12

which part

Answer 13

Plz give me full solution:)

Answer 14

maybe from the start...

\[x = \sqrt{3+\sqrt{3+ \sqrt{ 3+ \cdots}}}\]
\[x^2 = 3 + \sqrt{3+\sqrt{3 +\cdots}}\]
\[x^2 = 3+x\]
\[x^2 - x-3 = 0\]
you need to solve for x so use quadratic formula

Answer 15

better?

Answer 16

U mean \[\alpha={1+\sqrt{1-12} \over 2}, \beta={1-\sqrt{-11} \over 2} \]

Answer 17

-4(1)(-3) = +12

Answer 18

negative times negative = postive remember?

Answer 19

yes, but how we can gt \[2<x <3\]

Answer 20

\[\frac{1+\sqrt{13}}{2} = 2.3\]

Answer 21

try it in your calculator

Answer 22

\(\sqrt{13} = 3.6\)

Answer 23

lgba is right, this is concerned with quadratic formula.

Answer 24

1+ 3.6 = 4.6

4.6/2 = 2.3

Answer 25

I gt 3.16 (approximately).

Answer 26

sorry, 3.60

Answer 27

@lgbasallote

Answer 28

that's sqrt 13 right?

Answer 29

ya

Answer 30

so you have \[\frac{1+\sqrt{13}}{2} \implies \frac{1+3.6}{2} \implies \frac{4.6}{2} = 2.3\]

Answer 31

k!

Answer 32

\[x =\sqrt{3+\sqrt{3+\sqrt{3+\cdots}}}\]
\[x^2 = 3 + \sqrt{3+\sqrt{3+\cdots}}\]
\[x^2 = 3 + x\]
\[x^2 - x - 3 = 0\]
\[x = \frac{-(-1) \pm \sqrt{(-1) - 4(1)(-3)}}{2(1)}\]
\[x = \frac{1 \pm \sqrt{1 + 12}}{2}\]
\[x = \frac{1\pm \sqrt{13}}{2}\]
\[x = 2.3\]

Answer 33

K ! thanx a lot @lgbasallote