813° Вправа 813Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 243

813. Розкладіть на лінійні множники квадратний тричлен:

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

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

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

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

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

${\displaystyle x^2+3x+2=}$${\displaystyle \left(x+1\right)\left(x+2\right)}$;

2) ${\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 -9}$;

${\displaystyle x_2=}$${\displaystyle \frac{8-10}{-2}=}$${\displaystyle 1}$;

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

3) ${\displaystyle x^2-5x-24=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 1\cdot (-24)=}$${\displaystyle 25+96=}$${\displaystyle 121}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{121}}{2\cdot 1}=}$${\displaystyle \frac{5\pm 11}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+11}{2}=}$${\displaystyle 8}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-11}{2}=}$${\displaystyle -3}$;

${\displaystyle x^2-5x-24=}$${\displaystyle \left(x+3\right)\left(x-8\right)}$;

4) ${\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}$;

${\displaystyle x^2+6x+9=}$${\displaystyle {\left(x+3\right)}^2}$.