Question

$\displaystyle\sf x = 3+2\sqrt{2}$
$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$??
​

Answer 1

Answer:

Given

\sf\to x = 3+\sqrt{8}→x=3+

8

To find

\sf\to x^2+\dfrac{1}{x^2}→x

2

+

x

2

1

Now we can write as

\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}→

x

1

=

3+

8

1

Now rationalise

\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}\times\dfrac{3-\sqrt{8}}{3-\sqrt{8}}=\dfrac{3-\sqrt{8}}{9-8} =3-\sqrt{8}→

x

1

=

3+

8

1

×

3−

8

3−

8

=

9−8

3−

8

=3−

8

we get

\sf\to\dfrac{1}{x} =3-\sqrt{8}→

x

1

=3−

8

Using this identities

\to\sf (a+b)^2= a^2+b^2+2ab→(a+b)

2

=a

2

+b

2

+2ab

Now we can write as

\sf\to\bigg(x+\dfrac{1}{x}\bigg)^2=x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}→(x+

x

1

)

2

=x

2

+

x

2

1

+2×x×

x

1

Now put the value of x and 1/x

\sf\to (3+\sqrt{8}+3-\sqrt{8})^2=x^2+\dfrac{1}{x^2} +2→(3+

8

+3−

8

)

2

=x

2

+

x

2

1

+2

\sf\to (3+3)^2=x^2+\dfrac{1}{x^2} +2→(3+3)

2

=x

2

+

x

2

1

+2

\sf\to (6)^2=x^2+\dfrac{1}{x^2} +2→(6)

2

=x

2

+

x

2

1

+2

\sf\to 36=x^2+\dfrac{1}{x^2} +2→36=x

2

+

x

2

1

+2

\sf\to x^2+\dfrac{1}{x^2} =36-2→x

2

+

x

2

1

=36−2

\sf\to x^2+\dfrac{1}{x^2} =34→x

2

+

x

2

1

=34

Answer

\sf\to x^2+\dfrac{1}{x^2} =34→x

2

+

x

2

1

=34

Answer 2

$\huge\frak\red{Ꭺnsᗯer}$

$\begin{gathered} \sf \: given \: \: \boxed{ \sf u = 1 - {r}^{3} } \\ \ \\ \displaystyle \sf\int \limits \dfrac{ {9r}^{2} }{ \sqrt{1 - {r}^{3} } } \, dr \\ \\ \\ \sf \ \frac{d(1 - {r}^{3} )}{dr} = \frac{du}{dr} \\ \\ \\ \sf \: - 3 {r}^{3 - 1} = \frac{du}{dr} \\ \\ \\ \sf \: - 3 {r}^{2} = \frac{du}{dr} \\ \\ \\ \sf \: 3 \int \frac{ {3r}^{2} }{ \sqrt{1 - {r}^{3} } } dr \\ \\ \\ \sf \: - 3 \int{ \frac{ du}{ \sqrt{u} } } \\ \\ \\ \sf \: - 3 \int \: \frac{du}{ {u}^{ \frac{1}{2} } } \\ \\ \ \\ \sf \: - 3 \int( {u}^{ \frac{ - 1}{2}) } du \\ \\ \\ \sf \: - 3( \frac{ {u}^{? \frac{ - 1 + 2}{2} } }{ - \frac{1}{2} + 2} ) + c \\ \\ \\ \sf \: - 3( \frac{ {u}^{ \frac{1}{2} } }{ \frac{1}{2} } ) + c \\ \\ \\ \sf \: - 3(2 {u}^{ \frac{1}{2} } )+ c \\ \\ \\ \sf \: - 6 \sqrt{u} + c \\ \\ \sf \: put \: value \: of \: u : - \\ \\ \sf \: - 6 \sqrt{1 - {r}^{3} } + c\end{gathered}$