Question

ПОМОГИТЕ ПЖ! 2 ЗАДАНИЯ !

Answer 1

$\displaystyle\bf\\1)\\\\3x^{2} -12=0\\\\x^{2} -4=0\\\\(x-2)(x+2)=0\\\\x-2=0 \ \ \Rightarrow \ \ x_{1} =2\\\\x+2=0 \ \ \Rightarrow \ \ x_{2} =-2\\\\Otvet: \ - 2 \ ; \ 2\\\\\\2)\\\\3x^{2} +7x=0\\\\x(3x+7)=0\\\\x_{1} =0\\\\3x+7=0 \ \ 3x=-7 \ \ \Rightarrow \ \ x_{2} =-2\frac{1}{3} \\\\Otvet: \ 0 \ ; \ -2\frac{1}{3}$

$\displaystyle\bf\\1)\\\\(x-1)^{2} =2x-4\\\\x^{2}-2x+1-2x+4=0 \\\\x^{2} -4x+5=0\\\\D=(-4)^{2}-4\cdot 5=16 -20=-4 < 0$

Ответ : корней нет

$\displaystyle\bf\\2)\\\\\frac{1}{2} x^{2} -x-7=0\\\\x^{2} -2x-14=0\\\\D=(-2)^{2} -4\cdot(-14)=4+56=60=(2\sqrt{15})^{2} \\\\\\x_{1} =\frac{2-2\sqrt{15} }{2} =1-\sqrt{15} \\\\\\x_{2} =\frac{2+2\sqrt{15} }{2} =1+\sqrt{15} \\\\\\Otvet: \ 1-\sqrt{15} \ \ ; \ \ 1+\sqrt{15}$