Question

Найти коэффициенты уравнения х^2+рх+q=0 при условии, что разность корней уравнения равна 5, а разность их кубов равна 35

Answer 1

1)

{x1-x2 = 5

{x1³-x2³ = 35

{x1-x2 = 5

{(x1-x2)(x1²+x1•x2+x2²) = 35

{x1 = 5+x2

{x1²+x1•x2+x2² = 7

(5+x2)²+x2(5+x2)+x2²-7 = 0

25+10x2+x2²+5x2+x2²+x2²-7 = 0

3x2²+15x2+18 = 0

x2²+5x2+6 = 0

По Виету:

[x2 = -2

[x2 = -3

[x1 = 5+x2 = 3

[x1 = 5+x2 = 2

(3; -2), (2; -3)

2)По виету

{x1+x2 = -p =3-2 = 1

{x1•x2 = q = -6

x²-x-6 = 0

3)По Виету:

{-p = x1+x2 = 2-3 = -1

{x1•x2 = q = -6

x²+x-6 = 0

Answer 2

Ответ:

$x1 - x2 = 5 \\ {x}^{3} 1 - {x}^{3} 2 = 35 \\ (x1 - x2)( {x}^{2} 1 + x1x2 + {x}^{2} 2) = 35 \\ 5({x}^{2} 1 + x1x2 + {x}^{2} 2) = 35 \\ {x}^{2} 1 + x1x2 + {x}^{2} 2) = 7 \\ {(x1 - x2)}^{2} + 3x1x2 = 7 \\ 25 + 3x1x2 = 7 \\ 3x1x2 = - 18 \\ x1x2 = - 6 \\ x1x2 = q = - 6 \\ x1 = \frac{ - 6}{x2} \\ \frac{ - 6}{x2} - x2 = 5 \\ - 6 - {x}^{2} 2 = 5x2 \\ {x}^{2} 2 + 5x2 + 6 = 0 \\ x2 = - 3 \: \: \: \: \: x1 = \frac{ - 6}{ - 3} = 2\\ x2 = - 2 \: \: \: \: x1 = \frac{ - 6}{ - 2} = 3 \\ \\ x1 + x2 = - p1 = 2 + ( - 3) = - 1 = > p1 = 1 \\ x1 + x2 = - p2 = 3 + ( - 2) = 1 = > p 2= - 1$