Question

find real roots and difference of real roots of equation x=√6+x​

Answer 1

Answer:

${x}^{2} - x - \sqrt{6} = 0$

$x = \frac{1 + \sqrt{1 + 4 \sqrt{6} } }{2} , \frac{1 - \sqrt{1 + 4 \sqrt{6} } }{2}$

difference in the real roots is

$\frac{1 + \sqrt{1 + 4 \sqrt{6} } \: - 1 + \sqrt{1 + 4 \sqrt{6} } }{2}$

$= \frac{2 \sqrt{1 + 4 \sqrt{6} } }{2}$

$= \sqrt{1 + 4 \sqrt{6} }$