Question

if 2x=3+root 7, find the value of :4x square +1/x square​

Answer 1

The value of $4x^{2} +\frac{1}{x^{2} }$ is $\frac{256+96\sqrt{7} }{8+3\sqrt{7} }$.

Step-by-step explanation:

Given, 2x = 3 + $\sqrt{7}$

Squaring both sides,

$(2x)^{2} = (3+\sqrt{7} )^{2}$

$4x^{2} = 9 + 7 + 6\sqrt{7}$     $[(a+b)^{2} = a^{2} +b^{2} + 2ab]$

$4x^{2} = 16 + 6\sqrt{7}$ ................(i)

$x^{2} = \frac{(16+6\sqrt{7}) }{4}$...................(ii)

Now,

$4x^{2} + \frac{1}{x^{2} } \\= (16+6\sqrt{7} ) + \frac{1}{\frac{(16+6\sqrt{7}) }{4}}$                        [Using equation(i) & (ii)]

$= (16 + 6\sqrt{7} ) +\frac{4}{(16+6\sqrt{7}) } \\= \frac{(16+6\sqrt{7})^{2} +4 }{16+6\sqrt{7}} \\= \frac{256 + 252 + 192\sqrt{7}+4 }{16+6\sqrt7} \\= \frac{512+192\sqrt{7} }{16+6\sqrt{7}} \\=\frac{256+96\sqrt{7}}{8+3\sqrt{7}}$

Hence, the value of $4x^{2} +\frac{1}{x^{2} }$ is $\frac{256+96\sqrt{7} }{8+3\sqrt{7} }$.

Answer 2

Hope it helps you

Thanks for asking