Zadanie 1.
[tex]x\in \mathbb{R}[/tex]
[tex]x^4+x^2-12=0\\t=x^2\\t^2+t-12=0\\\Delta_t=1^2-4\cdot1\cdot12=1+48=49\\\sqrt{\Delta_t}=7\\\\\begin{array}{cc}t_1=\dfrac{-1-7}{2}=\dfrac{-8}2=-4&t_2=\dfrac{-1+7}2=\dfrac{6}2=3\end{array}[/tex]
[tex]x=\sqrt{t}\\\\x \neq \sqrt{-4}\\\boxed{x=\sqrt3 \vee x=-\sqrt3}[/tex]
Zadanie 2.
[tex]\left\{\begin{array}{c}y=x^2+2x-2\\y=-x^2-6x-2\end{array}\right\:\\\\x^2+2x-2=-x^2-6x-2\\x^2+x^2+2x+6x-2+2=0\\2x^2+8x=0 |:2\\x^2+4x=0\\x(x+4)=0\\\\\begin{array}{cc}x=0&x=-4\\y=0^2+2\cdot0-2&y=(-4)^2+2\cdot(-4)-2\\y=0+0-2&y=16-8-2\\y=-2&y=6\\(0; -2)&(-4; 6)\end{array}\\\\\boxed{(0;-2) \wedge (-4; 6)}[/tex]
Zadanie 3.
a)
[tex]15x^2+8x+1=0\\\Delta=64-60=4\\\sqrt{\Delta}=2\\x_1=\dfrac{-8-2}{30}=-\dfrac{10}{30}=-\dfrac13\\x_2=\dfrac{-8+2}{30}=-\dfrac{6}{30}=-\dfrac15[/tex]
Odp. Równanie ma dwa pierwiastki ujemne.
b)
[tex]2x^2-6x+4=0\\\Delta=36-32=4\\\sqrt{\Delta}=2\\x_1=\dfrac{6-2}{4}=\dfrac44=1\\x_2=\dfrac{6+2}4=\dfrac84=2[/tex]
Odp. Równanie ma dwa pierwiastki dodatnie.
