Question

If x = 9 + 4√5, find the value of x²+1/x²​

Answer 1

If x = 9 + 4√5, find the value of x²+1/x²See answers. Log in to add comment. Answer. 4.7/5. 224

Answer 2

$\star\:\:\bf\large\underline\blue{Question:-}$

• If $x = 9 + 4 \sqrt{5}$ find the value of ${x}^{2} + \frac{1}{ {x}^{2} }$

$\star\:\:\bf\large\underline\blue{Solution:-}$

$x = 9 + 4 \sqrt{5}$

$\implies \frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} }$

$= \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} }$

$= \frac{9 - 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5}) }^{2} }$

$= \frac{9 - 4 \sqrt{5} }{81 - 80}$

$= \frac{9 - 4 \sqrt{5} }{1}$

$= 9 - 4 \sqrt{5}$

Now,

$x + \frac{1}{x} = 9 + \cancel{ 4 \sqrt{5} } + 9 - \cancel{4 \sqrt{5} }$

$\implies x + \frac{1}{x} = 18$

Now, squaring both side, we get,

${(x + \frac{1}{x} )}^{2} = 81$

$\implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times \cancel{x} \times \cancel{x} = 81$

$\implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 81$

$\implies {x}^{2} + \frac{1}{ {x}^{2} } = 79$

$\star\:\:\bf\large\underline\blue{Answer:-}$

• ${x}^{2} + \frac{1}{ {x}^{2} } = 79$