1.
I sposob:
[tex]f(x)=(x-6)(x+3)\\f(x)=x^2+3x-6x-18\\f(x)=x^2-3x-18\\\Delta=(-3)^2-4*1*(-18)\\\Delta=9+72\\\Delta=81\\\sqrt{\Delta}=9\\x_1=\frac{3-9}2=\frac{-6}2=-3\\x_2=\frac{3+9}2=\frac{12}2=6[/tex]
II sposob:
[tex]\text{Odczytujemy miejsca zerowe z postaci iloczynowej}[/tex]
[tex]f(x)=a(x-x_1)(x-x_2)\\f(x)=a(x-6)(x+3)\\x_1=6\\x_2=-3[/tex]
2.
a)
[tex]x^2-4=0\\x^2=4\\x_1=\sqrt{4} = 2\\x_2=\sqrt{4}=-2[/tex]
b)
[tex]2x^2-16=0\\2x^2=16/:2\\x^2=8\\x_1=\sqrt{8}=2\sqrt{2}\\x_2=\sqrt{8}=-2\sqrt{2}[/tex]
c)
[tex]4-100x^2=0\\4=100x^2 /:100\\\frac{4}{100}=x^2\\x_1=\sqrt\frac{4}{100}=\frac{2}{10}=\frac15\\x_2=-\frac15[/tex]
3.
[tex]-x^2-5x-4<0\\\Delta=(-5)^2-4*(-1)*(-4)\\\Delta=25-16\\\Delta=9\\\sqrt{\Delta}=3\\x_1=\frac{5-3}{-2}=\frac{2}{-2}=-1\\x_2=\frac{5+3}{-2}=\frac8{-2}=-4\\a<0 \text{ - Ramiona paraboli skierowane w dol}\\\text{Rysujemy rysunek pogladowy paraboli i odczytujemy rozwiazania ponizej osi X}[/tex]
[tex]x\in(-\infty; -4)U(-1; \infty)[/tex]
1.
[tex]f(x) = (x-6)(x+3)\\\\(x-6)(x+3) = 0\\\\x-6 = 0 \ \vee \ x+3 = 0\\\\x = 6 \ \ \vee \ \ x = -3\\\\x \in \{-3,6\}[/tex]
2.
A)
[tex]x^{2}-4} = 0\\\\(x+2)(x-2) = 0\\\\x+2 = 0 \ \vee \ x-2 = 0\\\\x = -2 \ \ \vee \ \ x = 2\\\\x \in\{-2,2\}[/tex]
B)
[tex]2x^{2}-16 = 0\\\\2(x^{2}-8) = 0 \ \ /:2\\\\x^{2}-8 = 0\\\\(x+\sqrt{8})(x-\sqrt{8}) = 0\\\\(x + \sqrt{4\cdot2})(x - \sqrt{4\cdot2}) = 0\\\\(x + 2\sqrt{2})(x-2\sqrt{2}) = 0\\\\x+2\sqrt{2} = 0 \ \vee \ x - 2\sqrt{2} = 0\\\\x = -2\sqrt{2} \ \ \vee \ \ x = 2\sqrt{2}\\\\x \in \{-2\sqrt{2}, 2\sqrt{2}\}[/tex]
C)
[tex]4-100x^{2} = 0 \ \ /\cdot(-1)\\\\100x^{2}-4 = 0\\\\(10x+2)(10x-2) = 0\\\\10x+2 = 0 \ \vee \ 10x-2=0\\\\10x = -2 \ \vee \ 10x = 2\\\\x =-\frac{1}{5} \ \vee \ x = \frac{1}{5}\\\\x \in \{-\frac{1}{5}, \frac{1}{5}\}[/tex]
3.
[tex]-x^{2}-5x - 4 < 0 \ \ /\cdot (-1)\\\\x^{2}+5x+4 > 0\\\\\Delta = b^{2}-4ac=5^{2}-4\cdot1\cdot4 = 25-16 = 9\\\\\sqrt{\Delta} = \sqrt{9} = 3\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-5-3}{2\cdot1} = \frac{-8}{2} = -4\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-5+3}{2} = \frac{-2}{2} = -1\\\\x \in(-\infty; -4) \ \cup \ (-1;+\infty)[/tex]
