if x=5 -2X root3,find the value of x2+1/x2 and x4+1/x4 - Maths - Polynomials

Question

if x=5 -2X root3,find the value of x2+1/x2 and x4+1/x4

Ankush Jain · Accepted Answer

$Given, x = 5 - 2 \sqrt{3} \frac{1}{x} = \frac{1}{5 - 2 \sqrt{3}}$

On rationalizing the denominator on R.H.S, we get

$\frac{1}{x} = \frac{1}{5 - 2 \sqrt{3}} \times \frac{5 + 2 \sqrt{3}}{5 + 2 \sqrt{3}} = \frac{5 + 2 \sqrt{3}}{25 - 12} = \frac{5 + 2 \sqrt{3}}{13} {Now, x}^{2} = (5 - 2 \sqrt{3})^{2} = 25 + 12 - 20 \sqrt{3} = 37 - 20 \sqrt{3} and (\frac{1}{x})^{2} = {(\frac{5 + 2 \sqrt{3}}{13})}^{2} = \frac{1}{169} (25 + 12 + 20 \sqrt{3}) = \frac{37 + 20 \sqrt{3}}{169} \Rightarrow x^{2} + \frac{1}{x^{2}} = \frac{37 - 20 \sqrt{3} + 37 + 20 \sqrt{3}}{169} = \frac{6253 - 3380 \sqrt{3} + 37 + 20 \sqrt{3}}{169} = \frac{6290 - 3360 \sqrt{3}}{169}$

$Now, x^{4} + \frac{1}{x^{4}} = {(x^{2} + \frac{1}{x^{2}})}^{2} - 2 = (\frac{6290 - 3360 \sqrt{3}}{169})^{2} - 2 = \frac{39564100 + 33868800 - 42268800 \sqrt{3} - 57122}{28561} = \frac{7335778 - 42268800 \sqrt{3}}{28561}$

if x=5 -2X root3,find the value of x2+1/x2 and x4+1/x4.