1041 Вправа 1041Алгебра 8 клас Бевз 2025 · § 20 · стор. 197

1041. Склади квадратне рівняння з цілими коефіцієнтами, корені якого дорівнюють: а) 2 і 1/3; б) 3/4 і 2/3; в) 1/3 і −1 1/2; г) (3 − √7)/2 і (3 + √7)/2.

а) ${\displaystyle x_1=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle x_1+x_2=}$${\displaystyle \frac{7}{3}}$;

${\displaystyle x_1{x}_2=}$${\displaystyle \frac{2}{3}}$;

${\displaystyle x^2-\frac{7}{3}x+\frac{2}{3}=}$${\displaystyle 0}$ ${\displaystyle \mid \cdot 3}$;

${\displaystyle 3x^2-7x+2=}$${\displaystyle 0}$;

б) ${\displaystyle x_1=}$${\displaystyle \frac{3}{4}}$;

${\displaystyle x_2=}$${\displaystyle \frac{2}{3}}$;

${\displaystyle x_1+x_2=}$${\displaystyle \frac{9+8}{12}=}$${\displaystyle \frac{17}{12}}$;

${\displaystyle x_1{x}_2=}$${\displaystyle \frac{1}{2}}$;

${\displaystyle x^2-\frac{17}{12}x+\frac{1}{2}=}$${\displaystyle 0}$ ${\displaystyle \mid \cdot 12}$;

${\displaystyle 12x^2-17x+6=}$${\displaystyle 0}$;

в) ${\displaystyle x_1=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle -\frac{3}{2}}$;

${\displaystyle x_1+x_2=}$${\displaystyle \frac{2-9}{6}=}$${\displaystyle -\frac{7}{6}}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -\frac{1}{2}}$;

${\displaystyle x^2+\frac{7}{6}x-\frac{1}{2}=}$${\displaystyle 0}$ ${\displaystyle \mid \cdot 6}$;

${\displaystyle 6x^2+7x-3=}$${\displaystyle 0}$;

г) ${\displaystyle x_1=}$${\displaystyle \frac{3-\sqrt{7}}{2}}$;

${\displaystyle x_2=}$${\displaystyle \frac{3+\sqrt{7}}{2}}$;

${\displaystyle x_1+x_2=}$${\displaystyle \frac{6}{2}=}$${\displaystyle 3}$;

${\displaystyle x_1{x}_2=}$${\displaystyle \frac{9-7}{4}=}$${\displaystyle \frac{1}{2}}$;

${\displaystyle x^2-3x+\frac{1}{2}=}$${\displaystyle 0}$ ${\displaystyle \mid \cdot 2}$;

${\displaystyle 2x^2-6x+1=}$${\displaystyle 0}$.