978 Вправа 978Алгебра 8 клас Бевз 2025 · § 19 · стор. 184

978. Розв’яжи біквадратне рівняння. а) x⁴ − 5x² + 4 = 0; б) x⁴ − 17x² + 16 = 0; в) x⁴ − 3x² − 4 = 0; г) x⁴ + x² − 6 = 0; ґ) 4x⁴ − 3x² − 1 = 0; д) 3x⁴ − 7x² + 2 = 0.

Біквадратне рівняння ${\displaystyle a{x}^4+b{x}^2+c=}$${\displaystyle 0}$ розв’язуємо заміною ${\displaystyle x^2=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$.

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^4-17x^2+16=}$${\displaystyle 0}$;

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

${\displaystyle t^2-17t+16=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -17,c=}$${\displaystyle 16}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-17)}^2-4\cdot 1\cdot 16=}$${\displaystyle 289-64=}$${\displaystyle 225}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{17\pm \sqrt{225}}{2}=}$${\displaystyle \frac{17\pm 15}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{17+15}{2}=}$${\displaystyle 16}$;

${\displaystyle t_2=}$${\displaystyle \frac{17-15}{2}=}$${\displaystyle 1}$;

${\displaystyle x^2=}$${\displaystyle 16}$;

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

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

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

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

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

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-4)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{25}}{2}=}$${\displaystyle \frac{3\pm 5}{2}}$;

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

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

${\displaystyle t=}$${\displaystyle -1<0}$ — не підходить;

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

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

г) ${\displaystyle x^4+x^2-6=}$${\displaystyle 0}$;

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

${\displaystyle t^2+t-6=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 1,c=}$${\displaystyle -6}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 1^2-4\cdot 1\cdot (-6)=}$${\displaystyle 1+24=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle t_1=}$${\displaystyle \frac{-1+5}{2}=}$${\displaystyle 2}$;

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

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

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

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

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

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

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 4\cdot (-1)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{25}}{8}=}$${\displaystyle \frac{3\pm 5}{8}}$;

${\displaystyle t_1=}$${\displaystyle \frac{3+5}{8}=}$${\displaystyle 1}$;

${\displaystyle t_2=}$${\displaystyle \frac{3-5}{8}=}$${\displaystyle -0,25}$;

${\displaystyle t=}$${\displaystyle -0,25<0}$ — не підходить;

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

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

д) ${\displaystyle 3x^4-7x^2+2=}$${\displaystyle 0}$;

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

${\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 2}$;

${\displaystyle t_2=}$${\displaystyle \frac{7-5}{6}=}$${\displaystyle \frac{1}{3}}$;

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

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

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

${\displaystyle x=}$${\displaystyle \pm \sqrt{\frac{1}{3}}=}$${\displaystyle \pm \frac{\sqrt{3}}{3}}$.