1380 Вправа 1380Алгебра 8 клас Бевз 2025 · Повторення · стор. 263

1380. Розв’яжи рівняння. а) x² − 3x + 2 = 0; б) x² − 8x − 20 = 0; в) 4z² + z − 3 = 0; г) 3y² − 2y − 8 = 0; ґ) 0,25x² − 2x + 3 = 0; д) 2z² − 3z + 0,75 = 0.

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot 2=}$${\displaystyle 9-8=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{1}}{2}=}$${\displaystyle \frac{3\pm 1}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+1}{2}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{3-1}{2}=}$${\displaystyle 1}$;

б) ${\displaystyle x^2-8x-20=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -8,c=}$${\displaystyle -20}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-8)}^2-4\cdot 1\cdot (-20)=}$${\displaystyle 64+80=}$${\displaystyle 144}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{8\pm \sqrt{144}}{2}=}$${\displaystyle \frac{8\pm 12}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{8+12}{2}=}$${\displaystyle 10}$;

${\displaystyle x_2=}$${\displaystyle \frac{8-12}{2}=}$${\displaystyle -2}$;

в) ${\displaystyle 4z^2+z-3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle 1,c=}$${\displaystyle -3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 1^2-4\cdot 4\cdot (-3)=}$${\displaystyle 1+48=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{49}}{8}=}$${\displaystyle \frac{-1\pm 7}{8}}$;

${\displaystyle z_1=}$${\displaystyle \frac{-1+7}{8}=}$${\displaystyle 0,75}$;

${\displaystyle z_2=}$${\displaystyle \frac{-1-7}{8}=}$${\displaystyle -1}$;

г) ${\displaystyle 3y^2-2y-8=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -2,c=}$${\displaystyle -8}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-2)}^2-4\cdot 3\cdot (-8)=}$${\displaystyle 4+96=}$${\displaystyle 100}$;

${\displaystyle D>0}$, два корені:

${\displaystyle y_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{2\pm \sqrt{100}}{6}=}$${\displaystyle \frac{2\pm 10}{6}}$;

${\displaystyle y_1=}$${\displaystyle \frac{2+10}{6}=}$${\displaystyle 2}$;

${\displaystyle y_2=}$${\displaystyle \frac{2-10}{6}=}$${\displaystyle -\frac{4}{3}}$;

ґ) ${\displaystyle 0,25x^2-2x+3=}$${\displaystyle 0\mid \cdot 4}$;

${\displaystyle x^2-8x+12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -8,c=}$${\displaystyle 12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-8)}^2-4\cdot 1\cdot 12=}$${\displaystyle 64-48=}$${\displaystyle 16}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{8\pm \sqrt{16}}{2}=}$${\displaystyle \frac{8\pm 4}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{8+4}{2}=}$${\displaystyle 6}$;

${\displaystyle x_2=}$${\displaystyle \frac{8-4}{2}=}$${\displaystyle 2}$;

д) ${\displaystyle 2z^2-3z+0,75=}$${\displaystyle 0\mid \cdot 4}$;

${\displaystyle 8z^2-12z+3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 8,b=}$${\displaystyle -12,c=}$${\displaystyle 3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-12)}^2-4\cdot 8\cdot 3=}$${\displaystyle 144-96=}$${\displaystyle 48}$;

${\displaystyle D>0}$, два корені:

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{12\pm \sqrt{48}}{16}=}$${\displaystyle \frac{12\pm 4\sqrt{3}}{16}=}$${\displaystyle \frac{3\pm \sqrt{3}}{4}}$.