Question

-x²+x+12<0
Допоможіть

Answer 1

Ответ:

$\displaystyle \left( -\infty;-3 \right) \cup\left( 4 ;+\infty \right)}$

Объяснение:

$\displaystyle -x^{2} + x + 12 < 0$

Нулі:

$\displaystyle -x^{2} + x + 12 =0\\\\\\\\ a=-1 ,\ \ b=1 ,\ \ c=12\\\\ D = b^2 = 4ac = 1^2 = 4\cdot(- 1)\cdot12 = 1 + 48 = 49\\\\\\\sqrt{D} =\sqrt{49} = 7\\\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-1-7}{2\cdot(-1)}=\frac{-8 }{-2 }=4\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-1+7}{2\cdot(-1)}=\frac{6}{-2}=-3\\\\ \underline{x\in\left( -\infty;-3 \right) \cup\left( 4 ;+\infty \right)}$