Question

$4(x-\frac{1}{x}{)}^2-8(x+\frac{1}{8x})-29$ .Find the value of x.

Answer 1

$4(x-\frac{1}{x}{)}^2-8(x+\frac{1}{8x})-294(x^2-2+\frac{1}{x62})+8x+\frac{8}{x}=294x^2-8+\frac{4}{12}+8x+\frac{8}{x}=294x^2-8x^2+4+8x^3+8x=29x^{2}4x^2-8x^3-37x^2+8x+4=0=P(0\left)P\right(2)=4(2{)}^4+8(2{)}^3-37(2{)}^2+8\left(2\right)+4=64+64-148+16+4=128-148+20=-20+20=0P\left(\frac{1}{2}\right)=4\left(\frac{1}{16}\right)+8\left(\frac{1}{8}\right)-\frac{37}{4}+\frac{8}{2}+4=\frac{1}{4}+1-\frac{37}{4}+4+4=9-\frac{36}{4}=9-9=0(x-2)\&(x-\frac{1}{2})allthefactors(x-2)(x-\frac{1}{2})=x^2-\frac{5}{2}x+1$


$4x^2+18x+4=02x^2+9x+2=0x=-9\pm \frac{\sqrt{9^2-4\left(2\right)\left(2\right)}}{2\left(2\right)}=-\frac{9\pm \sqrt{81-16}}{4}=-\frac{9\pm \sqrt{65}}{4}\therefore x=2,\frac{1}{2},-\frac{9\pm \sqrt{81-16}}{4},-\frac{9-\sqrt{81-16}}{4}$