Question

Реши уравнения:
2) x² + 7x + 12 = 0;
8) x² - 61 = 0.​

Answer 1

Ответ:

2) $-4,\ -3$

8) $\displaystyle-\sqrt{61},\ \sqrt{61}$

Объяснение:

2)

$\displaystyle x^{2} + 7 x + 12 =0\\\\ a=1 ,\ \ b=7 ,\ \ c=12\\\\ D = b^2 - 4ac = 7^2 - 4\cdot1\cdot12 = 49 - 48 = 1\\\\\sqrt{D} =\sqrt{1} = 1\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-7-1}{2\cdot1}=\frac{-8 }{2 }=-4\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-7+1}{2\cdot1}=\frac{-6}{2}=-3$

8)

$\displaystylex^{2} -61=0\\\\(x-\sqrt{61})(x+\sqrt{61})=0\\\\x-\sqrt{61}=0\ \ \ \ \ \ \ \ \ \ x+\sqrt{61}=0\\\\x=\sqrt{61}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-\sqrt{61}$