How to use Vieta's formula? Or is there Simpler way to do?
What i know is $x_1 + x_2 = \frac{-b}{a}$, $x_1 . x_2 = \frac{c}{a}$
2026-03-25 12:53:59.1774443239
$x^4-6x^3+(13-m^2)x^2 - 12x+4$
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Viet theorem says:
$${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }=6\\ { x }_{ 1 }{ x }_{ 2 }+{ x }_{ 1 }{ x }_{ 3 }+{ x }_{ 1 }{ x }_{ 4 }+{ x }_{ 2 }{ x }_{ 3 }+{ x }_{ 2 }{ x }_{ 4 }+{ x }_{ 3 }{ x }_{ 4 }=13-{ m }^{ 2 }\\ { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }+{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 4 }+{ x }_{ 1 }{ x }_{ 3 }{ x }_{ 4 }+{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 }=12\\ { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 }=4$$
so we get $$ \\ \frac { 1 }{ { x }_{ 1 }^{ 2 } } +\frac { 1 }{ { x }_{ 2 }^{ 2 } } +\frac { 1 }{ { x }_{ 3 }^{ 2 } } +\frac { 1 }{ { x }_{ 4 }^{ 2 } } =\frac { { \left( { x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 } \right) }^{ 2 }+{ \left( { x }_{ 1 }{ x }_{ 3 }{ x }_{ 4 } \right) }^{ 2 }+{ \left( { x }_{ 1 }{ x }_{ 2 }{ x }_{ 4 } \right) }^{ 2 }+{ \left( { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } \right) }^{ 2 } }{ { \left( { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 } \right) }^{ 2 } } =\\ \\ =\frac { { \left( { x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 }+{ x }_{ 1 }{ x }_{ 3 }{ x }_{ 4 }+{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 4 }+{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } \right) }^{ 2 } }{ { \left( { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 } \right) }^{ 2 } } -\frac { 2\left( { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }^{ 2 }{ x }_{ 4 }^{ 2 }+{ x }_{ 1 }{ x }_{ 3 }{ x }_{ 2 }^{ 2 }{ x }_{ 4 }^{ 2 }+{ x }_{ 1 }{ x }_{ 4 }{ x }_{ 2 }^{ 2 }{ x }_{ 3 }^{ 2 }+{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 1 }^{ 2 }{ x }_{ 4 }^{ 2 }+{ x }_{ 2 }{ x }_{ 4 }{ x }_{ 1 }^{ 2 }{ x }_{ 3 }^{ 2 }+{ x }_{ 3 }{ x }_{ 4 }{ x }_{ 1 }^{ 2 }{ x }_{ 2 }^{ 2 } \right) }{ { \left( { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 } \right) }^{ 2 } } \\ \\ \\ =\frac { { \left( { x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 }+{ x }_{ 1 }{ x }_{ 3 }{ x }_{ 4 }+{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 4 }+{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } \right) }^{ 2 }-2{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 }\left( { x }_{ 1 }{ x }_{ 2 }+{ x }_{ 1 }{ x }_{ 3 }+{ x }_{ 1 }{ x }_{ 4 }+{ x }_{ 2 }{ x }_{ 3 }+{ x }_{ 2 }{ x }_{ 4 }+{ x }_{ 3 }{ x }_{ 4 } \right) }{ { \left( { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 } \right) }^{ 2 } } =\\ \\ =\frac { { 12 }^{ 2 }-8\left( 13-{ m }^{ 2 } \right) }{ { 4 }^{ 2 } } =\frac { 40+8{ m }^{ 2 } }{ 16 } =\frac { 5 }{ 2 } +\frac { { m }^{ 2 } }{ 2 } $$