Question

Rozwiąż nierówności
a) x²-25 > 0
b) -3x²+2x ≥ 0
c) -5x²+x-7 < 0

Answer 1

Odpowiedź:

[tex]\huge\boxed{a)~x^{2} -25~ > 0~\Leftrightarrow~x\in(-\infty:-5)\cup(5;+\infty)}[/tex]

[tex]\huge\boxed{b)~~-3x^{2} +2x\geq 0~~\Leftrightarrow~~x\in\left < 0;\dfrac{2}{3}\right > ~~}[/tex]

[tex]\huge\boxed{c)~~-5x^{2} +x-7 < 0~~\Leftrightarrow~~x\in \mathbb{R}~~}[/tex]

Szczegółowe wyjaśnienie:

Korzystamy ze wzorów:

• [tex]y=ax^{2} +bx+c\\\\\Delta =b^{2} -4ac\\\\Gdy~~\Delta > 0~~\Rightarrow~~x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\\\Gdy~~\Delta =0~~\Rightarrow~~x_{o} =\dfrac{-b}{2a} \\\\Gdy~~\Delta < 0~~\Rightarrow~~brak~~pierwiastkow[/tex]

• [tex]Gdy~~a > 0~~\Rightarrow~~ramiona ~~paraboli~~skierowane~~do~~gory\\\\Gdy~~a < 0~~\Rightarrow~~ramiona ~~paraboli~~skierowane~~w~~dol[/tex]

• [tex]x^{2} -y^{2} =(x-y)\cdot (x+y)[/tex]
• [tex](-)\cdot (-)=(+)[/tex]
• [tex](+)\cdot (-)=(-)[/tex]

Rozwiązujemy:

[tex]a)\\\\x^{2} -25 > \\\\a=1,~~b=0,~~c=-25\\\\x^{2} -5^{2} > 0\\\\(x-5)(x+5) > 0\\\\x-5=0~~\lor~~x+5=0\\\\x=5~~\lor~~x=-5\\\\a=1~~\Rightarrow~~ramiona ~~paraboli~~skierowane~~do~~gory\\\\\\\huge\boxed{~~x^{2} -25~ > 0~~\Leftrightarrow~~x\in(-\infty:-5)\cup(5;+\infty)~~}[/tex]

[tex]b)\\\\-3x^{2} +2x\geq 0\\\\a=-3,~~b=2,~~c=0\\\\x(-3x+2)\geq 0\\\\x=0~~\lor~~-3x+2=0\\\\x=0~~\lor~~x=\dfrac{2}{3} \\\\a=-3~~\Leftrightarrow~~ramiona~~paraboli~~skierowane~~w~~dol\\\\\\\huge\boxed{~~-3x^{2} +2x\geq 0~~\Leftrightarrow~~x\in\left < 0;\dfrac{2}{3}\right > ~~}[/tex]

[tex]c)\\\\-5x^{2} +x-7 < 0\\\\a=-5,~~b=1,~~c=-7\\\\\Delta=1^{1}-4\cdot (-5)\cdot (-7)\\\\\Delta=1-140\\\\\Delta=-139\\\\a=-5~~\Rightarrow~~ramiona~~paraboli~~skierowane~~w~~dol\\\\\huge\boxed{~~-5x^{2} +x-7 < 0~~\Leftrightarrow~~x\in \mathbb{R}~~}[/tex]