1027 Вправа 1027Алгебра 8 клас Бевз 2025 · § 20 · стор. 196

1027. Розв’яжи рівняння. а) y² + 5y − 14 = 0; б) z² − 2z − 15 = 0; в) c² − 2c − 8 = 0; г) t² + 9t − 10 = 0.

а) ${\displaystyle y^2+5y-14=}$${\displaystyle 0}$;

${\displaystyle y_1+y_2=}$${\displaystyle -5}$;

${\displaystyle y_1{y}_2=}$${\displaystyle -14}$;

підбираємо: ${\displaystyle -7+2=}$${\displaystyle -5}$;

${\displaystyle -7\cdot 2=}$${\displaystyle -14}$;

${\displaystyle y_1=}$${\displaystyle -7}$;

${\displaystyle y_2=}$${\displaystyle 2}$;

б) ${\displaystyle z^2-2z-15=}$${\displaystyle 0}$;

${\displaystyle z_1+z_2=}$${\displaystyle 2}$;

${\displaystyle z_1{z}_2=}$${\displaystyle -15}$;

підбираємо: ${\displaystyle 5+(-3)=}$${\displaystyle 2}$;

${\displaystyle 5\cdot (-3)=}$${\displaystyle -15}$;

${\displaystyle z_1=}$${\displaystyle 5}$;

${\displaystyle z_2=}$${\displaystyle -3}$;

в) ${\displaystyle c^2-2c-8=}$${\displaystyle 0}$;

${\displaystyle c_1+c_2=}$${\displaystyle 2}$;

${\displaystyle c_1{c}_2=}$${\displaystyle -8}$;

підбираємо: ${\displaystyle 4+(-2)=}$${\displaystyle 2}$;

${\displaystyle 4\cdot (-2)=}$${\displaystyle -8}$;

${\displaystyle c_1=}$${\displaystyle 4}$;

${\displaystyle c_2=}$${\displaystyle -2}$;

г) ${\displaystyle t^2+9t-10=}$${\displaystyle 0}$;

${\displaystyle t_1+t_2=}$${\displaystyle -9}$;

${\displaystyle t_1{t}_2=}$${\displaystyle -10}$;

підбираємо: ${\displaystyle -10+1=}$${\displaystyle -9}$;

${\displaystyle -10\cdot 1=}$${\displaystyle -10}$;

${\displaystyle t_1=}$${\displaystyle -10}$;

${\displaystyle t_2=}$${\displaystyle 1}$.