993 Вправа 993Алгебра 8 клас Бевз 2025 · § 19 · стор. 186

993. Не виконуючи побудови, знайди координати точок перетину графіків функцій. а) y = x² + x − 8 і y = −2x² + 5x − 9; б) y = 15/(x + 1) і y = x + 3.

а) ${\displaystyle x^2+x-8=}$${\displaystyle -2x^2+5x-9}$;

${\displaystyle x^2+x-8+2x^2-5x+9=}$${\displaystyle 0}$;

${\displaystyle 3x^2-4x+1=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -4,c=}$${\displaystyle 1}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-4)}^2-4\cdot 3\cdot 1=}$${\displaystyle 16-12=}$${\displaystyle 4}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{4\pm \sqrt{4}}{6}=}$${\displaystyle \frac{4\pm 2}{6}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{4-2}{6}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle y\left(1\right)=}$${\displaystyle 1+1-8=}$${\displaystyle -6}$;

${\displaystyle y\left(\frac{1}{3}\right)=}$${\displaystyle \frac{1}{9}+\frac{1}{3}-8=}$${\displaystyle \frac{1+3-72}{9}=}$${\displaystyle -\frac{68}{9}=}$${\displaystyle -7\frac{5}{9}}$;

Точки: ${\displaystyle \left(1;-6\right)}$, ${\displaystyle \left(\frac{1}{3};-7\frac{5}{9}\right)}$;

б) ${\displaystyle \frac{15}{x+1}=}$${\displaystyle x+3}$, ${\displaystyle x\ne -1}$;

${\displaystyle 15=}$${\displaystyle (x+3)(x+1)}$;

${\displaystyle 15=}$${\displaystyle x^2+4x+3}$;

${\displaystyle 0=}$${\displaystyle x^2+4x-12}$;

${\displaystyle x^2+4x-12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 4,c=}$${\displaystyle -12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 4^2-4\cdot 1\cdot (-12)=}$${\displaystyle 16+48=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-4\pm \sqrt{64}}{2}=}$${\displaystyle \frac{-4\pm 8}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-4+8}{2}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{-4-8}{2}=}$${\displaystyle -6}$;

${\displaystyle y\left(2\right)=}$${\displaystyle 2+3=}$${\displaystyle 5}$;

${\displaystyle y(-6)=}$${\displaystyle -6+3=}$${\displaystyle -3}$;

Точки: ${\displaystyle \left(2;5\right)}$, ${\displaystyle \left(-6;-3\right)}$.