Question

If x⁴+1/x⁴= 194 find x+1/x?

Answer 1

$HEYA \: \\ \\ let \: \: \: \: (x + 1 \div x) = z \\ \\ squaring \: on \: both \: sides \: we \: have \: \\ \\ (x + 1 \div x) {}^{2} = z {}^{2} \\ \\ x {}^{2} + 1 \div x {}^{2} = z {}^{2} - 2....Equation \: \: (i) \\ \\ squaring \: the \: above \: equation \: on \: both \: sides \: we \: have \: \\ \\ (x {}^{2} + 1 \div x {}^{2} ) {}^{2} = (z {}^{2} - 2) {}^{2} \\ \\ x {}^{4} + 1 \div x {}^{4} = (z {}^{2} - 2) {}^{2} - 2 \\ \\ 194 + 2 = (z {}^{2} - 2) {}^{2} \\ \\ (z {}^{2} - 2) {}^{2} = 196 \\ \\ (z {}^{2} - 2) = \sqrt{196} \\ \\ (z {}^{2} - 2) = + 14 \\ or \\ (z {}^{2} - 2) = - 14 \\ \\ z {}^{2} = 16 \\ or \\ z {}^{2} = - 12(complex \: roots \: ) \\ \\ z = 4 \: \: \: or \: \: \: z = - 4 \\ \\ so \: \: \: \: x + 1 \div x = 4 \\ or \\ x + 1 \div x = - 4 \\ \\ if \: u \: are \: in \: above \: 10th \: class \: then \: two \ \\ more \: values \: are \: also \: possible \: \\ i.e \\ \\ x + 1 \div x = 2 \sqrt{3} i \\ or \\ x + 1 \div x = - 2 \sqrt{3} i \\ where \: i \: is \: iota$