967 Вправа 967Алгебра 8 клас Бевз 2025 · § 19 · стор. 183

967. Розв’яжи рівняння, що зводиться до квадратного. а) (2x − 3)² = 8x; б) (2x + 1)² = 3x + 4; в) 2(3z + 9) = (3z + 5)²; г) 12(3 − x) = (3x − 1)².

Квадрат двочлена: ${\displaystyle {(a\pm b)}^2=}$${\displaystyle a^2\pm 2ab+b^2}$.

а) ${\displaystyle {(2x-3)}^2=}$${\displaystyle 8x}$;

${\displaystyle 4x^2-12x+9=}$${\displaystyle 8x}$;

${\displaystyle 4x^2-12x+9-8x=}$${\displaystyle 0}$;

${\displaystyle 4x^2-20x+9=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle -20,c=}$${\displaystyle 9}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-20)}^2-4\cdot 4\cdot 9=}$${\displaystyle 400-144=}$${\displaystyle 256}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{20\pm \sqrt{256}}{8}=}$${\displaystyle \frac{20\pm 16}{8}}$;

${\displaystyle x_1=}$${\displaystyle \frac{20+16}{8}=}$${\displaystyle 4,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{20-16}{8}=}$${\displaystyle 0,5}$;

б) ${\displaystyle {(2x+1)}^2=}$${\displaystyle 3x+4}$;

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

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

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 1^2-4\cdot 4\cdot (-3)=}$${\displaystyle 1+48=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{49}}{8}=}$${\displaystyle \frac{-1\pm 7}{8}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+7}{8}=}$${\displaystyle 0,75}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-7}{8}=}$${\displaystyle -1}$;

в) ${\displaystyle 2(3z+9)=}$${\displaystyle {(3z+5)}^2}$;

${\displaystyle 6z+18=}$${\displaystyle 9z^2+30z+25}$;

${\displaystyle 0=}$${\displaystyle 9z^2+30z+25-6z-18}$;

${\displaystyle 9z^2+24z+7=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 9,b=}$${\displaystyle 24,c=}$${\displaystyle 7}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 24^2-4\cdot 9\cdot 7=}$${\displaystyle 576-252=}$${\displaystyle 324}$;

${\displaystyle D>0}$, два корені:

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-24\pm \sqrt{324}}{18}=}$${\displaystyle \frac{-24\pm 18}{18}}$;

${\displaystyle z_1=}$${\displaystyle \frac{-24+18}{18}=}$${\displaystyle -\frac{1}{3}}$;

${\displaystyle z_2=}$${\displaystyle \frac{-24-18}{18}=}$${\displaystyle -\frac{7}{3}}$;

г) ${\displaystyle 12(3-x)=}$${\displaystyle {(3x-1)}^2}$;

${\displaystyle 36-12x=}$${\displaystyle 9x^2-6x+1}$;

${\displaystyle 0=}$${\displaystyle 9x^2-6x+1-36+12x}$;

${\displaystyle 9x^2+6x-35=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 9,b=}$${\displaystyle 6,c=}$${\displaystyle -35}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 6^2-4\cdot 9\cdot (-35)=}$${\displaystyle 36+1260=}$${\displaystyle 1296}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-6\pm \sqrt{1296}}{18}=}$${\displaystyle \frac{-6\pm 36}{18}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-6+36}{18}=}$${\displaystyle \frac{5}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-6-36}{18}=}$${\displaystyle -\frac{7}{3}}$.