Question

if x=√5-2/√5+2and y=√5+2/5-2findx²
​

Answer 1

Given:−

x = \sqrt{5} + 2x=

5

+2

\underline{\bold{To\:find:-}}

Tofind:−

\begin{gathered} {x}^{2} + \frac{1}{ {x}^{2} } \\ \end{gathered}

x 2+x

21\underline{\bold{Solution:-}}

Solution:−

\begin{gathered}x = \sqrt{5} + 2 \\ \\ \frac{1}{x} = \frac{1}{ \sqrt{5} + 2 } \\ \\ \bold{On \: rationalising \: them} \\ \\ \frac{1}{x} = \frac{1}{ \sqrt{5} + 2 } \times \frac{ \sqrt{5} - 2}{ \sqrt{5} - 2}\\ \end{gathered}

x= 5 +2x1

= 5 +21

Onrationalisingthem

x1 = 5 +21 × 5 −25 −2

\begin{gathered}\bold{Using \: identity } \\ \end{gathered}

Usingidentity

\begin{gathered}\bold {{a}^{2} - {b}^{2} = (a + b)(a - b)}\\ \\ \frac{1}{x} = \frac{ \sqrt{5} - 2 }{ { (\sqrt{5}) }^{2} - {2}^{2} } \\ \\ \frac{1}{x} = \frac{ \sqrt{5} - 2 }{5 - 4} \\ \\ \frac{1}{x} = \sqrt{5} - 2 \\ \end{gathered}

a 2 −b 2

=(a+b)(a−b)x1

= (5) 2−225−2x1

= 5−45−2x1

= 5−2

Now,

\begin{gathered}x + \frac{1}{x} = \sqrt{5} + 2 + \sqrt{5} - 2 \\ \\ x + \frac{1}{x} = 2 \sqrt{5} \\ \\ \bold{On \: squaring \: both \: sides} \\ \\ {(x + \frac{1}{x}) }^{2} = {(2 \sqrt{5} )}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 4 \times 5 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 20 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 20 - 2 \\ \\\boxed{\bold{ {x}^{2} + \frac{1}{ {x}^{2} } = 18}}\end{gathered}

x+ x1= 5+2+ 5−2x+ x1

=25

Onsquaringbothsides

(x+x1) 2=(2 5) 2x 2+ x 21+2×x× x1

=4×5x2+ x 21+2=20x 2+ x21

=20−2x2+ x 21

=18