1075 Вправа 1075Алгебра 8 клас Бевз 2025 · § 21 · стор. 203

1075. Знайди корені квадратного тричлена. а) x² + 8x − 9; б) 2x² − 5x − 7; в) 5x² + 2x − 3; г) y² − y − 6; ґ) 4z² − 5z + 1; д) 3n² − n − 2.

Корені тричлена — корені рівняння ${\displaystyle a{x}^2+bx+c=}$${\displaystyle 0}$.

а) ${\displaystyle x^2+8x-9}$: ${\displaystyle x_1+x_2=}$${\displaystyle -8}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -9}$ → ${\displaystyle x_1=}$${\displaystyle 1,x_2=}$${\displaystyle -9}$;

б) ${\displaystyle 2x^2-5x-7}$:

${\displaystyle 2x^2-5x-7=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -5,c=}$${\displaystyle -7}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot (-7)=}$${\displaystyle 25+56=}$${\displaystyle 81}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{5\pm \sqrt{81}}{4}=}$${\displaystyle \frac{5\pm 9}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+9}{4}=}$${\displaystyle 3,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-9}{4}=}$${\displaystyle -1}$;

в) ${\displaystyle 5x^2+2x-3}$:

${\displaystyle 5x^2+2x-3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 5,b=}$${\displaystyle 2,c=}$${\displaystyle -3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 2^2-4\cdot 5\cdot (-3)=}$${\displaystyle 4+60=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-2\pm \sqrt{64}}{10}=}$${\displaystyle \frac{-2\pm 8}{10}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-2+8}{10}=}$${\displaystyle 0,6}$;

${\displaystyle x_2=}$${\displaystyle \frac{-2-8}{10}=}$${\displaystyle -1}$;

г) ${\displaystyle y^2-y-6}$: ${\displaystyle y_1+y_2=}$${\displaystyle 1}$;

${\displaystyle y_1{y}_2=}$${\displaystyle -6}$ → ${\displaystyle y_1=}$${\displaystyle 3,y_2=}$${\displaystyle -2}$;

ґ) ${\displaystyle 4z^2-5z+1}$:

${\displaystyle 4z^2-5z+1=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle -5,c=}$${\displaystyle 1}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-5)}^2-4\cdot 4\cdot 1=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{5\pm \sqrt{9}}{8}=}$${\displaystyle \frac{5\pm 3}{8}}$;

${\displaystyle z_1=}$${\displaystyle \frac{5+3}{8}=}$${\displaystyle 1}$;

${\displaystyle z_2=}$${\displaystyle \frac{5-3}{8}=}$${\displaystyle 0,25}$;

д) ${\displaystyle 3n^2-n-2}$:

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

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -1,c=}$${\displaystyle -2}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 3\cdot (-2)=}$${\displaystyle 1+24=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle n_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{25}}{6}=}$${\displaystyle \frac{1\pm 5}{6}}$;

${\displaystyle n_1=}$${\displaystyle \frac{1+5}{6}=}$${\displaystyle 1}$;

${\displaystyle n_2=}$${\displaystyle \frac{1-5}{6}=}$${\displaystyle -\frac{2}{3}}$.