rozwiąz mi to ktos na szybko plska

[tex]Zad. 1\\\\x^2+6x+2=0\\\Delta=6^2-4*1*2=36-8=28\\\sqrt{\Delta}=\sqrt{28}=2\sqrt7\\\\x_1=\frac{-6-2\sqrt7}2=\frac{2(-3-\sqrt7)}2=-3-\sqrt7\\x_2=\frac{-6+2\sqrt7}2=\frac{2(-3+\sqrt7)}2=-3+\sqrt7\\\\x_1+x_2=-3-\sqrt7+(-3+\sqrt7)=-3-\sqrt7-3+\sqrt7=-6[/tex]

[tex]Zad. 2\\2x^2+bx+3=0\\\Delta > 0\\\Delta=b^2-4*2*3 > 0\\\Delta=b^2-24 > \\b^2-24 > 0\\b^2 > 24\\b > \sqrt{24}\\b > 2\sqrt{6} \text{ v } b < 2\sqrt6\\\\b\in R / \{2\sqrt6\}[/tex]

[tex]b\in(-\infty; 2\sqrt6)U(2\sqrt6; \infty)[/tex]

Zad. 3

[tex]f(x)=-2x^2-3x+1\\a < 0 - \text{ ramiona skierowane w dol}\\\text{Jezeli wierzcholek znajduje sie w przedziale, jest on wartoscia najwieksza}\\p=\frac{3}{-4}=-\frac34 \in < -3; 0 > \\f_{max}=\frac{-\Delta}{4a}=\frac{-((-3)^2-4*(-2)*1)}{-8}=\frac{-(9+8)}{-8}=\frac{17}8[/tex]

Zad. 4

[tex]\left \{ {{x+y=1} \atop {y=-(x-2)^2+1}} \right. \\\left \{ {{y=1-x} \atop {1-x=-(x^2-4x+4)+1}} \right. \\\left \{ {{y=1-x} \atop {1-x=-x^2+4x-4+1}} \right. \\\left \{ {{y=1-x} \atop {x^2-4x+3+1-x=0}} \right. \\\left \{ {{y=1-x} \atop {x^2-5x+4=0}} \right. \\\Delta=25-16=9\\\sqrt{\Delta}=3\\x_1=\frac{5-3}2=\frac22=1\\x_2=\frac{5+3}2=\frac82=4\\\\y_1=1-1=0\\y_2=1-4=-3\\\\A(1, 0), B(4, -3)[/tex]

Basetlago Basetlago · Answer 2

1.

[tex]x^{2}+6x+2 = 0\\\\a = 1, \ b = 6, \ c = 2\\\\Z \ wzorow \ Viete'a\\\\x_1+x_2 = \frac{-b}{a}, \ \ \ \ x_1\cdot x_2 = \frac{c}{a}\\\\\Delta = b^{2}-4ac = 6^{2}-4\cdot1\cdot2 = 36-8 = 28 > 0\\\\x_1+x_2 = \frac{-b}{a} = \frac{-6}{1} =\boxed{ -6}[/tex]

2.

[tex]2x^{2}+6x+3 = 0\\\\a = 2, \ b = 6, \ c = 3\\\\\Delta = b^{2}-4ac = 6^{2}-4\cdot2\cdot3 = 36-24 = 12 > 0\\\\x_1\cdot x_2 = \frac{c}{a} = \frac{3}{2} > 0\\\\x_1+x_2 = \frac{-b}{a} = \frac{-6}{2} = -3[/tex]

Stąd wniosek, że równanie ma pierwiastki ujemne.

3.

[tex]f(x) = -2x^{2}-3x+1, \ \ \ \ \langle-3;0\rangle\\\\a = -2, \ b = -3, \ c = 1\\\\p = \frac{-b}{2a} = \frac{-(-3)}{2\cdot(-2)} =\frac{3}{-4} = -\frac{3}{4} \ \ \in \langle-3;0\rangle\\\\f(-3) = -2\cdot(-3)^{2}-3\cdot(-3) + 1 = -2\cdot9+9+1 = -18+10 =\boxed{ -8}\\\\f(0) = -2\cdot0^{2}-3\cdot0+1 = 1\\\\q = f(p) = f(-\frac{3}{4}) = -2\cdot(-\frac{3}{4})^{2}-3\cdot(-\frac{3}{4})+1 = -2\cdot\frac{9}{16}+\frac{9}{4}+1 = -\frac{9}{8}+\frac{18}{8}+\frac{8}{8} = \frac{17}{8}[/tex]

W przedziale < -3;0 > funkcja f przyjmuje najmniejszą wartość równą -8.

4.

[tex]x+y = 1\\y = -(x-2)^{2}+1\\\\y = 1-x\\y =-(x^{2}-4x+4)+1 = -x^{2}+4x-4+1 = -x^{2}+4x-3\\\\1-x = -x^{2}+4x-3\\\\x^{2}-x-4x+1+3 = 0\\\\x^{2}-5x+4 = 0\\\\a = 1, \ b = -5, \ c = 4\\\\\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{2}{2} = 1\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{5+3}{2} = \frac{8}{2} = 4[/tex]

[tex]y_1 = 1-1 = 0\\\\y_2 = 1-4 = -3\\\\\\\{x_1 = 1\\\{y_1 = 0\\\\\{x_2 = 4\\\{y_2 = -3[/tex]