Question

plz answer these question​

Answer 1

AnSwEr

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

x= 3 + 2√2

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

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

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

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

x=3+2√2

Now 1/x

• Rationalize it

$\tt\rightarrow{\dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}}$

$\tt\rightarrow{\dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}\times\dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}}$

=>1/x = 3-2√2 / 9-8

=>1/x = 3-2√2 / 1

Now adding

=>x + 1/x =3-2√2+3+2√2

=>x + 1/x =6

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

$\tt{X+\dfrac{1}{X}}$=6

Both side squaring

$\tt{(X+\dfrac{1}{X})^2}$=(6)²

$\implies\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}=36$

$\implies\tt{36=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

$\implies\tt{36=X^2+\dfrac{1}{X^2}+2}$

$\implies\tt{36-2=X^2+\dfrac{1}{X^2}}$

$\implies\tt{34=X^2+\dfrac{1}{X^2}}$

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

$\implies\tt{X^3+\dfrac{1}{X^3}={6}(34-1)}$

$\implies\tt{X^3+\dfrac{1}{X^3}={6}(33)}$

$\implies\tt{X^3+\dfrac{1}{X^3}={198}}$

Now

$\tt{\sqrt{x}+\dfrac{1}{\sqrt{x}}}$

Both side squaring

$\tt{\sqrt{x}+\dfrac{1}{\sqrt{x}}}$=x + 1/x + 2

=>√x + 1/√x = 6 +2

=>√x + 1/√x = 8