982 Вправа 982Алгебра 8 клас Бевз 2025 · § 19 · стор. 185

982. Використовуючи метод заміни змінної, розв’яжи рівняння. а) x − 8√x + 15 = 0; б) x + 3√x − 10 = 0; в) 2x − 5√x + 2 = 0; г) 3x + 7√x + 2 = 0.

а) ${\displaystyle x-8\sqrt{x}+15=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

${\displaystyle t^2-8t+15=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -8,c=}$${\displaystyle 15}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-8)}^2-4\cdot 1\cdot 15=}$${\displaystyle 64-60=}$${\displaystyle 4}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle t_1=}$${\displaystyle \frac{8+2}{2}=}$${\displaystyle 5}$;

${\displaystyle t_2=}$${\displaystyle \frac{8-2}{2}=}$${\displaystyle 3}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 5}$;

${\displaystyle x=}$${\displaystyle 25}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 3}$;

${\displaystyle x=}$${\displaystyle 9}$;

б) ${\displaystyle x+3\sqrt{x}-10=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

${\displaystyle t^2+3t-10=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 3,c=}$${\displaystyle -10}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 1\cdot (-10)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-3\pm \sqrt{49}}{2}=}$${\displaystyle \frac{-3\pm 7}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-3+7}{2}=}$${\displaystyle 2}$;

${\displaystyle t_2=}$${\displaystyle \frac{-3-7}{2}=}$${\displaystyle -5}$;

${\displaystyle t=}$${\displaystyle -5<0}$ — не підходить;

${\displaystyle \sqrt{x}=}$${\displaystyle 2}$;

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

в) ${\displaystyle 2x-5\sqrt{x}+2=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

${\displaystyle 2t^2-5t+2=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -5,c=}$${\displaystyle 2}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot 2=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{5\pm \sqrt{9}}{4}=}$${\displaystyle \frac{5\pm 3}{4}}$;

${\displaystyle t_1=}$${\displaystyle \frac{5+3}{4}=}$${\displaystyle 2}$;

${\displaystyle t_2=}$${\displaystyle \frac{5-3}{4}=}$${\displaystyle 0,5}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 2}$;

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

${\displaystyle \sqrt{x}=}$${\displaystyle 0,5}$;

${\displaystyle x=}$${\displaystyle 0,25}$;

г) ${\displaystyle 3x+7\sqrt{x}+2=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

${\displaystyle 3t^2+7t+2=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle 7,c=}$${\displaystyle 2}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 7^2-4\cdot 3\cdot 2=}$${\displaystyle 49-24=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-7\pm \sqrt{25}}{6}=}$${\displaystyle \frac{-7\pm 5}{6}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-7+5}{6}=}$${\displaystyle -\frac{1}{3}}$;

${\displaystyle t_2=}$${\displaystyle \frac{-7-5}{6}=}$${\displaystyle -2}$;

обидва значення ${\displaystyle t<0}$ — коренів немає.