Question

If x = 2+√3 then 1/x

Answer 1

Answer:

ko

Step-by-step explanation:

have you got answer thanks for support

Answer 2

AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• x = 2+√3

${\sf{\green{\underline{\large{To\:find}}}}}$

• 1/x

${\sf{\pink{\underline{\Large{Explanation}}}}}$

X= 2+√3

Then

1/x = 1/{2+√3}

We have 2 rationalize it

1/x = 1/{2+√3}

=>1/x = 1/{2+√3} × {2-√3} /{2-√3}

=>1/x = {2-√3}/1

Additional Information

=>x + 1/x = 2-√3 + 2+√3

=>x + 1/x =4

We know that

$\boxed{\boxed{\sf\red{(a+b)^2=a^2+b^2+2ab}}}$

Therefore

$\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

Solving

$\tt{X+\dfrac{1}{X}}$=4

Both side squaring

$\tt{(X+\dfrac{1}{X})^2}$=(4)²

$\implies\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}=16$

$\implies\tt{16=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

$\implies\tt{16=X^2+\dfrac{1}{X^2}+2}$

$\implies\tt{16-2=X^2+\dfrac{1}{X^2}}$

$\implies\tt{14=X^2+\dfrac{1}{X^2}}$

Now

$\tt{X^3+\dfrac{1}{X^3}}$

Formula used

$\implies\tt{X^3+\dfrac{1}{X^3)}=(X+\dfrac{1}{x})(X^2+\dfrac{1}{X^2}-\frac{1}{X}\times{X)}}$

¶utting value

$\implies\tt{X^3+\dfrac{1}{X^3}={4}(14-1)}$

$\implies\tt{X^3+\dfrac{1}{X^3}={4}(13)}$

$\implies\tt{X^3+\dfrac{1}{X^3}={52}}$