Question

Minimum integral value of a for
which the equation no
x^4 - 2ax²+x+a^2 - a=0 has all
its roots real.​

Answer 1

$\bf \huge \blue{Answer : - }$

$\sf \: given \: \: the \: \: equation \: \\ \\ \sf {x}^{4} - 2ax + x + {a}^{2} - a = 0 \\ \\ \sf \: considering \: \: it \: \: is \: \: a \: \: quadratic \: \: equation \: \: in \: a \: {a}^{2} - a(1 + {2x}^{2} ) + {x}^{4} + x = 0 \\ \\ \sf \: solving \: \: the \: \: equation \: \: we \: \: get \\ \\ \sf \: \frac{a = - [ - (1 + {2x}^{2} )] ± \sqrt{ - (1 + 2x) - 4( {x}^{4} + x)} }{2} \\ \\ \sf \: = \frac{(1 + {2x}^{2} )± \sqrt{1 + {4x}^{2} + {4x}^{4} - {4x}^{4} - 4x} }{2} \\ \\ \sf \: = \frac{(1 + {2x}^{2}) ±(2x - 1)}{2} \\ \\ \sf \therefore \: a = \frac{1 + {2x}^{2} + 2x - 1 }{2} \: \: and \: \: \frac{1 + {2x}^{2} - 2x - 1 }{2} \\ \\ a = {x}^{2} + x \: \: \: and \: \: \: {x}^{2} - x + 1 \\ \\ ths \: \: we \: \: get \: \: a \: \: fact$

$( {x}^{2} + x - a)( {x}^{2} - x + 1 - a) = 0 \\ \\ for \: \: {x}^{2} + x - a = 0 \\ \\ x = \frac{ - 1± \sqrt{1 + 4a} }{2} \\ \\ for \: \: the \: \: roots \: \: to \: \: be \: \: real \: \: 1 + 4a \ge \: 0 \\ \\ = a \ge \: \frac{ - 1}{4} \\ \\ so \: \: a \: \: has \: \: the \: \: roots \: \: in \: \left [\frac{ - 1}{4} ,\infty \right ] \\ \\ for \: \: {x}^{2} - x + 1 - a = 0 \\ \\ x = \frac{ - 1± \sqrt{4a - 3} }{2} \\ \\ for \: \: the \: \: roots \: \: to \: \: be \: \: real \\ \\ 4a - 3 \ge0 \\ \\ a \ge \frac{3}{4}$

$so \: \: a \: \: has \: \: the \: \: roots \: \: in \: \: [ \frac{3}{4} , \infty )$