36 Вправа 36 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 320

36. Розв’яжіть квадратне рівняння:

1) ${\displaystyle x^2+8x-9=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 8^2-4\cdot 1\cdot (-9)=}$${\displaystyle 64+36=}$${\displaystyle 100}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-8\pm \sqrt{100}}{2\cdot 1}=}$${\displaystyle \frac{-8\pm 10}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-8+10}{2}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-8-10}{2}=}$${\displaystyle -9}$;

2) ${\displaystyle x^2-6x-7=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-6)}^2-4\cdot 1\cdot (-7)=}$${\displaystyle 36+28=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{6\pm \sqrt{64}}{2\cdot 1}=}$${\displaystyle \frac{6\pm 8}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{6+8}{2}=}$${\displaystyle 7}$;

${\displaystyle x_2=}$${\displaystyle \frac{6-8}{2}=}$${\displaystyle -1}$;

3) ${\displaystyle x^2-6x+9=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-6)}^2-4\cdot 1\cdot 9=}$${\displaystyle 36-36=}$${\displaystyle 0}$;

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

4) ${\displaystyle x^2+6x+5=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 6^2-4\cdot 1\cdot 5=}$${\displaystyle 36-20=}$${\displaystyle 16}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-6\pm \sqrt{16}}{2\cdot 1}=}$${\displaystyle \frac{-6\pm 4}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-6+4}{2}=}$${\displaystyle -1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-6-4}{2}=}$${\displaystyle -5}$.