1047 Вправа 1047Алгебра 8 клас Бевз 2025 · § 20 · стор. 197

1047. Знайди корені рівняння x1 і x2 (x2 > x1) і значення k, якщо: а) x² + kx + 10 = 0 і x1 − x2 = −3; б) x² − 8x + k = 0 і x1 : x2 = −0,2.

а) ${\displaystyle x_2=}$${\displaystyle x_1+3}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 10}$;

нехай ${\displaystyle x_1=}$${\displaystyle t}$:

${\displaystyle t(t+3)=}$${\displaystyle 10}$;

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 3,c=}$${\displaystyle -10}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 1\cdot (-10)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-3\pm \sqrt{49}}{2}=}$${\displaystyle \frac{-3\pm 7}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-3+7}{2}=}$${\displaystyle 2}$;

${\displaystyle t_2=}$${\displaystyle \frac{-3-7}{2}=}$${\displaystyle -5}$;

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

${\displaystyle x_2=}$${\displaystyle 5}$;

${\displaystyle k=}$${\displaystyle -(2+5)=}$${\displaystyle -7}$;

2) ${\displaystyle x_1=}$${\displaystyle -5}$;

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

${\displaystyle k=}$${\displaystyle -(-5-2)=}$${\displaystyle 7}$;

б) ${\displaystyle x_1=}$${\displaystyle -0,2x_2}$;

${\displaystyle x_1+x_2=}$${\displaystyle 8}$;

${\displaystyle -0,2x_2+x_2=}$${\displaystyle 8}$;

${\displaystyle 0,8x_2=}$${\displaystyle 8}$;

${\displaystyle x_2=}$${\displaystyle 10}$;

${\displaystyle x_1=}$${\displaystyle -0,2\cdot 10=}$${\displaystyle -2}$;

${\displaystyle k=}$${\displaystyle x_1{x}_2=}$${\displaystyle -2\cdot 10=}$${\displaystyle -20}$.