[tex]\bold{Zad. 14}\\w(x)=(x^2+4x-5)(2x^2+7)\\x^2+4x-5 = 0\\\Delta=4^2-4*1*(-5)=16+20=36\\\sqrt{\Delta}=6\\x_1=\frac{-4-6}2=\frac{-10}2=-5\\x_2=\frac{-4+6}2=\frac22=1\\\\2x^2+7=0 /-7\\\Delta=0^2-4*2*7=-56 - \text{brak miejsc zerowych}\\\\\underline{Odp. A}[/tex]
[tex]\bold{Zad. 15}\\\\\frac{2x^3-3x^2-20x+21}{x+3}=2x^2-9x+7\\\\2x^2-9x+7=0\\\Delta=(-9)^2-4*2*7=81-56=25\\\sqrt{\Delta}=5\\x_1=\frac{9-5}4=\frac44=1\\x_2=\frac{9+5}4=\frac{14}4=\frac72=3.5\\\\x+3=0 /-3\\x=-3\\\\\underline{Odp. A}[/tex]
[tex]\bold{Zad. 16}\\\\(x^2-x-6)(x^2+4x+4)=0\\\\x^2-x-6=0\\\Delta=(-1)^2-4*1*(-6)=1+24=25\\\sqrt{\Delta}=5\\x_1=\frac{1-5}2=\frac{-4}2=-2\\x_2=\frac{1+5}2=\frac62=3\\\\x^2+4x+4=0\\\Delta=4^2-4*1*4=16-16=0\\x_3=\frac{-4}{2}=-2\\\\x_1+x_2+x_3=-2+3-2=1-2=-1\\\\\underline{Odp. D}[/tex]
[tex]\bold{Zad. 17}\\\\x^3+4x^2-2x-8=0\\x^2(x+4)-2(x+4)=0\\(x^2-2)(x+4)=0\\(x-\sqrt2)(x+\sqrt2)(x+4)=\\\\x_1-\sqrt2=0 /+\sqrt2\\x_1=\sqrt2\\\\x_2+\sqrt2=0 /-\sqrt2\\x_2=-\sqrt2\\\\x_3+4=0 /-4\\x_3=-4\\\\\underline{Odp. A}[/tex]
[tex]\bold{Zad. 18}[/tex]
[tex](x+3)(x-2)^3(x-3)^5 > 0\\(x+3)(x-2)(x-2)^2(x-3)(x-3)^2(x-3)^2 > 0\\x+3 > 0 /-3\\x > -3\\x-2 > 0 /+2\\x > 2\\(x-2)^2 > 0\\x^2-4x+4 > 0\\\Delta=(-4)^2-4*1*4=16-16=0\\x_0=\frac{4}2=2\\x\in(-\infty; 2)U(2; \infty)\\x-3 > 0 /+3\\x > 3\\(x-3)^2 > 0\\x^2-6x+9 > 0\\\Delta=(-6)^2-4*1*9=36-36=0\\x_0=\frac{6}2=3\\x\in(-\infty; 3)U(3; \infty)\\\\\underline{x\in(\infty; -3)U(3; \infty)}[/tex]
