Please help me to find this questions answer x/2> -1 + 3x/4 ; x element N
Answers
GiVeN : -
x = 3 + 2√2
To FiNd : -
\sqrt{x} + \frac{1}{ \sqrt{x} }
x
+
x
1
SoLuTiOn : -
Here,
x = 3 + 2√2. ,
Then,
\begin{gathered} \implies \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ \implies \: \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \implies \: \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2)} }^{2} } \\ \\ \implies \: \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} = 3 - 2 \sqrt{2} \end{gathered}
⟹
x
1
=
3+2
2
1
⟹
x
1
=
3+2
2
1
×
3−2
2
3−2
2
⟹
x
1
=
(3)
2
−(2
2)
2
3−2
2
⟹
x
1
=
9−8
3−2
2
=3−2
2
Now,
We have
=> x = 3 + 2√2 and 1/x = 3 - 2√2
Now,
We know that,
\implies \: {( \sqrt{x} + \frac{1}{ \sqrt{x} } ) }^{2} = x + \frac{1}{x} + 2⟹(
x
+
x
1
)
2
=x+
x
1
+2
So,
Put the above values in the above identity
We get,
\begin{gathered} \implies \: {( \sqrt{x} + \frac{1}{ \sqrt{x} } ) }^{2} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} + 2 \\ \\ \implies \: {( \sqrt{x} + \frac{1}{ \sqrt{x} } )}^{2} =8 \\ \\ \implies \: \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{8} = 2 \sqrt{2} \end{gathered}
⟹(
x
+
x
1
)
2
=3+2
2
+3−2
2
+2
⟹(
x
+
x
1
)
2
=8
⟹
x
+
x
1
=
8
=2
2
So, we get,
=> √x + 1/√x = 2√2