Prove the following ?

Question

$x + \frac{1}{x} = 4$ $x+1/x=4$
$x = \sqrt{7 + \sqrt{3}}$ $x=sqrt(7+sqrt3)$

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Answer 1 · 2018-06-25T04:29:26

$x = 2 \pm \sqrt{3}$ $x=2+-sqrt3$

$x + \frac{1}{x} = 4$ $x+1/x=4$ where $x \neq 0$ $x !=0$

$x^{2} + 1 = 4 x$ $x^2+1=4x$

$x^{2} - 4 x + 1 = 0$ $x^2-4x+1=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{4 \pm \sqrt{16 - 4}}{2}$ $x=(4+-sqrt(16-4))/2$

$x = \frac{4 \pm \sqrt{12}}{2}$ $x=(4+-sqrt12)/2$

$x = \frac{4 \pm 2 \sqrt{3}}{2}$ $x=(4+-2sqrt3)/2$

$x = 2 \pm \sqrt{3}$ $x=2+-sqrt3$

Answer 2 · 2018-06-25T05:16:34

Given

$x+1/x=4.....(1)$ where $x !=0$

$(x-1/x)^2=(x+1/x)^2-4*x*1/x$

$=>(x-1/x)^2=4^2-4=12$

$=>x-1/x=pm2sqrt3....(2)$

Adding (1) and (2)

$2x=4pm2sqrt3$

$=>x=2pmsqrt3$