Question

Розв’язати рівняння: (2х²+х+1)×(2х²+х+3)=8

Answer 1

Ответ:

Объяснение:

$(2x^{2} +x+1)(2x^{2} +x+3)=8\\2x^{2} +x=t\\(t+1)(t+3)=8\\t^2+4t-5=0\\(t+5)(t-1)=0\\t_{1} =-5 \\t_{2}=1\\2x^2+x+5=0 D=b^2-4ac=1-4*2*5=-39 < 0\\2x^2+x-1=0 D=b^2-4ac=1-4*2*(-1)=9\\x_{1}=\frac{-1-3}{4}=-1\\ x_{2}=\frac{-1+3}{4}=\frac{1}{2}$

Answer 2

$\displaystyle\bf\\(2x^{2} +x+1)(2x^{2} +x+3)=8\\\\2x^{2} +x+2=m \ \ \Rightarrow \ \ 2x^{2} +x+1=m-1 \ \ ; \ \ 2x^{2} +x+3=m+1\\\\\\(m-1)\cdot(m+1)=8\\\\m^{2} -1=8\\\\m^{2} =9\\\\m_{1} =-3 \ \ ; \ \ m_{2} =3\\\\1)\\\\2x^{2} +x+2=-3\\\\2x^{2} +x+5=0\\\\D=1^{2} -4\cdot 2\cdot 5=1-40=-39 < 0\\\\x\in \oslash\\\\2)\\\\2x^{2} +x+2=3\\\\2x^{2} +x-1=0\\\\D=1^{2} -4\cdot 2\cdot(-1)=1+8=9=3^{2} \\\\x_{1}=\frac{-1-3}{4} =-1\\\\\\x_{2} =\frac{-1+3}{4} =\frac{1}{2} =0,5$

$\displaystyle\bf\\Otvet \ : \ -1 \ ; \ 0,5$