Question

if x=5-2sqrt(6) then find the value x^2+1/x^2

Answer 1

@amistre64 ....... can u pls.... check this out

Answer 2

@ajprincess ....... can u pls.... check this out

Answer 3

put 5-2sqrt(6) in place of x and find the value

Answer 4

im getting a little a troubled in between can u pls...do the problem with every step....pls....

Answer 5

we would perfer not to "do all the work"; but to be able to guide you thru the process, interacting with youalong the way

Answer 6

is this\[x^2+\frac1{x^2}\]or\[\frac{x^2+1}{x^2}\]

Answer 7

im sry for the trouble but 2morrow is my exzam..and i dnt hv much time,,

Answer 8

its ..\[x ^{2}+\frac{ 1 }{ x ^{2} }\]

Answer 9

\[(x+\frac{1}{x})^2\]
\[=x^2+2*x*\frac{1}{x}+\frac{1}{x^2}\]
\[=x^2+2+\frac{1}{x^2}\]
\[(x+\frac{1}{x})^2-2=x^2+\frac{1}{x^2}\]
Nw put 5-2sqrt(6) for x in
\[(x+\frac{1}{x})^2-2
and try to find the value.

Answer 10

Nw put 5-2sqrt(6) for x in
\[(x+\frac{1}{x})^2-2\] and try to find the value.

Answer 11

and \(x=5-2\sqrt{6}\) so every place we see an "x" we plug in its value, right?

Answer 12

i think aj might have misread the notation. which is common when trying to decipher the ascii text

Answer 13

since this is the form you picked
\[x^2+\frac1{x^2}\]lets make life simpler by working the general setup first; then we can insert the specifics

\[x^2\frac{x^2}{x^2}+\frac1{x^2}\]
\[\frac{x^4+1}{x^2}\]
\[\frac{(5-2\sqrt{6})^4+1}{(5-2\sqrt{6})^2}\]

Answer 14

your other problems gave you approximations for sqrts; does this one include approximations for sqrt(6) ??

Answer 15

@amistre64 , i think aj's method works just fine and is simpler, can u plz check it again ??

Answer 16

let me see :)

Answer 17

http://www.wolframalpha.com/input/?i=x%5E2%2B1%2Fx%5E2+-+%28%28x%2B1%2Fx%29%5E2-2%29

i agree

Answer 18

aj apparently read it correctly, good job :)

Answer 19

Thanx a lot for the compliment:)

Answer 20

Can u find the value @basith?

Answer 21

giv me a min....

Answer 22

ha k.

Answer 23

ill do it here step by step ....pls see if it is correct...ok @ajprincess

Answer 24

\[(5-2\sqrt{6}+\frac{ 1 }{ 5-\sqrt{6} })^{2}-2\]

Answer 25

\[\frac{ 25-20\sqrt{6}+24 }{ 5+2\sqrt{6} }+\frac{ 1 }{ 25-20\sqrt{6}+24 }\]

Answer 26

then wat????????? and it is nt 1 it is\[5+2\sqrt{6}\]

Answer 27

\[(5-2\sqrt6+\frac{ 1 }{ 5-2\sqrt6 })^2-2\]
\[=(\frac{ 25-20\sqrt6+24 }{5-2\sqrt6}+\frac{ 1 }{ 5-2\sqrt6})^2-2\]
\[=(\frac{ 25-20\sqrt6+24 + 1 }{ 5-2\sqrt6 })^2-2\]

Answer 28

Can u continue from this @basith?

Answer 29

????......wat 2 do next..??? :(

Answer 30

@ajprincess

Answer 31

Simplify the numerator and factorise it

Answer 32

\[(\frac{ 50-20\sqrt{6} }{ 5-2\sqrt{6} })^{2}-2\]

Answer 33

yup right.
Nw try to factorise the numerator.

Answer 34

is that = \[(10)^{2}-2\] @ajprincess

Answer 35

ya:)

Answer 36

so...98

Answer 37

yup:)

Answer 38

i hv 1 mor question....pls wait

Answer 39

if x+1/x=7 , then find the value of x^3+1/x^3

Answer 40

i sthe answer just 343

Answer 41

@ajprincess

Answer 42

May I knw hw u got that answer?

Answer 43

\[(x+\frac{ 1 }{ x })^{3}=(7)^{3}=343\]

Answer 44

it is really wrong....isn't it.. :(

Answer 45

hmm ya.
\[x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2\]
\[x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2-1+\frac{1}{x^2})\]
\[x^3+\frac{1}{x^3}=(x+\frac{1}{x})((x+\frac{1}{x})^2-2-1)\]
\[=(x+\frac{1}{x})((x+\frac{1}{x})^2-2-1)\]
\[=(x+\frac{1}{x})((x+\frac{1}{x})^2-3)\]
Hope u can do it nw.

Answer 46

wat did u di in the 3rd step???

Answer 47

can u plz type it down?

Answer 48

@basith a simpler way to do that will be to use:
\(\large (a+b)^3=a^3+b^3+3ab(a+b)\)
here put a= x and b = 1/x
so ab = x* (1/x) = 1
a+b = x+1/x = 7
can u solve further ??

Answer 49

actually wat did u write first.....\[x ^{3}+\frac{ 1 }{ x ^{3} }=(x+\frac{ 1 }{ ^{3} })^{2}-2\] ????????????

Answer 50

@ajprincess