1386 Вправа 1386Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1386. Розв’яжи рівняння. а) x⁴ − 5x² + 4 = 0; б) x⁴ − 10x² + 9 = 0.

а) ${\displaystyle x^4-5x^2+4=}$${\displaystyle 0}$;

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

${\displaystyle t^2-5t+4=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -5,c=}$${\displaystyle 4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-5)}^2-4\cdot 1\cdot 4=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

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

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

${\displaystyle t_2=}$${\displaystyle \frac{5-3}{2}=}$${\displaystyle 1}$;

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

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

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

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

б) ${\displaystyle x^4-10x^2+9=}$${\displaystyle 0}$;

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

${\displaystyle t^2-10t+9=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -10,c=}$${\displaystyle 9}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-10)}^2-4\cdot 1\cdot 9=}$${\displaystyle 100-36=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{10\pm \sqrt{64}}{2}=}$${\displaystyle \frac{10\pm 8}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{10+8}{2}=}$${\displaystyle 9}$;

${\displaystyle t_2=}$${\displaystyle \frac{10-8}{2}=}$${\displaystyle 1}$;

${\displaystyle x^2=}$${\displaystyle 9}$;

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

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

${\displaystyle x=}$${\displaystyle \pm 1}$.