1387 Вправа 1387Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1387. Розв’яжи рівняння. а) 3x⁴ − 2x² − 40 = 0; б) 5y⁴ + 7y² − 12 = 0.

а) ${\displaystyle 3x^4-2x^2-40=}$${\displaystyle 0}$;

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

${\displaystyle 3t^2-2t-40=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -2,c=}$${\displaystyle -40}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-2)}^2-4\cdot 3\cdot (-40)=}$${\displaystyle 4+480=}$${\displaystyle 484}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{2\pm \sqrt{484}}{6}=}$${\displaystyle \frac{2\pm 22}{6}}$;

${\displaystyle t_1=}$${\displaystyle \frac{2+22}{6}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{2-22}{6}=}$${\displaystyle -\frac{10}{3}}$;

${\displaystyle x^2=}$${\displaystyle 4}$;

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

${\displaystyle t=}$${\displaystyle -\frac{10}{3}<0}$ — не підходить;

б) ${\displaystyle 5y^4+7y^2-12=}$${\displaystyle 0}$; заміна ${\displaystyle y^2=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

${\displaystyle 5t^2+7t-12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 5,b=}$${\displaystyle 7,c=}$${\displaystyle -12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 7^2-4\cdot 5\cdot (-12)=}$${\displaystyle 49+240=}$${\displaystyle 289}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-7\pm \sqrt{289}}{10}=}$${\displaystyle \frac{-7\pm 17}{10}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-7+17}{10}=}$${\displaystyle 1}$;

${\displaystyle t_2=}$${\displaystyle \frac{-7-17}{10}=}$${\displaystyle -2,4}$;

${\displaystyle y^2=}$${\displaystyle 1}$;

${\displaystyle y=}$${\displaystyle \pm 1}$;

${\displaystyle t=}$${\displaystyle -2,4<0}$ — не підходить.