793° Вправа 793Алгебра 8 клас Тарасенкова 2025 · § 19 · стор. 236

793. Складіть квадратне рівняння, коренями якого є числа:

Для ${\displaystyle x^2+bx+c=}$${\displaystyle 0}$: ${\displaystyle b=}$${\displaystyle -(x_1+x_2)}$, ${\displaystyle c=}$${\displaystyle x_1\cdot x_2}$.

1) ${\displaystyle 4}$ і ${\displaystyle 3}$;

${\displaystyle x_1+x_2=}$${\displaystyle 4+3=}$${\displaystyle 7}$; ${\displaystyle b=}$${\displaystyle -7}$;

${\displaystyle x_1\cdot x_2=}$${\displaystyle 4\cdot 3=}$${\displaystyle 12}$; ${\displaystyle c=}$${\displaystyle 12}$;

${\displaystyle x^2-7x+12=}$${\displaystyle 0}$;

2) ${\displaystyle -2}$ і ${\displaystyle 5}$;

${\displaystyle x_1+x_2=}$${\displaystyle -2+5=}$${\displaystyle 3}$; ${\displaystyle b=}$${\displaystyle -3}$;

${\displaystyle x_1\cdot x_2=}$${\displaystyle -2\cdot 5=}$${\displaystyle -10}$; ${\displaystyle c=}$${\displaystyle -10}$;

${\displaystyle x^2-3x-10=}$${\displaystyle 0}$;

3) ${\displaystyle -4}$ і ${\displaystyle \frac{3}{4}}$;

${\displaystyle x_1+x_2=}$${\displaystyle -4+\frac{3}{4}=}$${\displaystyle -3,25=}$${\displaystyle -\frac{13}{4}}$; ${\displaystyle b=}$${\displaystyle \frac{13}{4}}$;

${\displaystyle x_1\cdot x_2=}$${\displaystyle -4\cdot \frac{3}{4}=}$${\displaystyle -3}$; ${\displaystyle c=}$${\displaystyle -3}$;

${\displaystyle x^2+\frac{13}{4}x-3=}$${\displaystyle 0}$ | ${\displaystyle \cdot 4}$

${\displaystyle 4x^2+13x-12=}$${\displaystyle 0}$;

4) ${\displaystyle -2}$ і ${\displaystyle -\frac{3}{8}}$;

${\displaystyle x_1+x_2=}$${\displaystyle -2+(-\frac{3}{8})=}$${\displaystyle -2,375=}$${\displaystyle -\frac{19}{8}}$; ${\displaystyle b=}$${\displaystyle \frac{19}{8}}$;

${\displaystyle x_1\cdot x_2=}$${\displaystyle -2\cdot (-\frac{3}{8})=}$${\displaystyle 0,75}$; ${\displaystyle c=}$${\displaystyle 0,75}$;

${\displaystyle x^2+\frac{19}{8}x+\frac{3}{4}=}$${\displaystyle 0}$ | ${\displaystyle \cdot 8}$

${\displaystyle 8x^2+19x+6=}$${\displaystyle 0}$.