1008 Вправа 1008Алгебра 8 клас Бевз 2025 · § 19 · стор. 187

1008. Розв’яжи рівняння. а) x² − (a + 5)x + 3a + 6 = 0; б) ax² − (2a + 1)x + a + 1 = 0.

а) ${\displaystyle D=}$${\displaystyle {(a+5)}^2-4(3a+6)}$=;

=${\displaystyle a^2+10a+25-12a-24=}$${\displaystyle a^2-2a+1=}$${\displaystyle {(a-1)}^2}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{a+5\pm (a-1)}{2}}$;

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

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

якщо ${\displaystyle a=}$${\displaystyle 1}$, то корінь один: ${\displaystyle x=}$${\displaystyle 3}$;

б) якщо ${\displaystyle a=}$${\displaystyle 0}$: ${\displaystyle -x+1=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 1}$;

якщо ${\displaystyle a\ne 0}$: ${\displaystyle D=}$${\displaystyle {(2a+1)}^2-4a(a+1)}$=;

=${\displaystyle 4a^2+4a+1-4a^2-4a=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{2a+1\pm 1}{2a}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{2a}{2a}=}$${\displaystyle 1}$.

Відповідь: при ${\displaystyle a=}$${\displaystyle 0}$ ${\displaystyle x=}$${\displaystyle 1}$; при ${\displaystyle a\ne 0}$ ${\displaystyle x=}$${\displaystyle 1}$ або ${\displaystyle x=}$${\displaystyle \frac{a+1}{a}}$.