Задание 2, стр 60

Question

Решите уравнение.
$$a)3 x^{2}-7 x+4=0;\quad c)5 x^{2}+8 x+3=0$$
$$b)5 x^{2}+6 x+1=0;\quad d)2 x^{2}-9 x+7=0$$

Gimli · Accepted Answer

\begin{enumerate}[label=\alph*)]
\item $3 x^{2}-7 x+4=0$
$$D=b^{2}-4 a c=(-7)^{2}-3 \cdot 4 \cdot 4=49-48=1=1^{2} .$$ $$x_{1}=\frac{-b+\sqrt{D}}{2 a}=\frac{7+1}{6}=\frac{8}{6}=1 \frac{1}{3} ; x_{2}=\frac{-b-\sqrt{D}}{2 a}=\frac{7-1}{6}=1$$ 
	extbf{OTBET}: $x \in\left\{1 \frac{1}{3}: 1ight\} .$
\item  $5 x^{2}+6 x+1=0$
$$D=b^{2}-4 a c=6^{2}-4 \cdot 5 \cdot 1=36-20=16=4^{2}$$
$$x_{1}=\frac{-b+\sqrt{D}}{2 a}=\frac{-6+4}{10}=-\frac{2}{10}=-\frac{1}{5} ; x_{2}=\frac{-b-\sqrt{D}}{2 a}=\frac{-6-4}{10}=-1$$
	extbf{OTBET:} $x \in\left\{-\frac{1}{5} ;-1ight\}$
\item $5 x^{2}+8 x+3=0$
$$D=b^{2}-4 a c=8^{2}-5 \cdot 3 \cdot 4=64-60=4=2^{2}$$
$$x_{1}=\frac{-b+\sqrt{D}}{2 a}=\frac{-8+2}{10}=-\frac{6}{10}=-\frac{3}{5} ; x_{2}=\frac{-b-\sqrt{D}}{2 a}=\frac{-8-2}{10}=-1$$
	extbf{OTBET:} $x \in\left\{-\frac{3}{5} ;-1ight\}$
\item $2 x^{2}-9 x+7=0$
$$D=b^{2}-4 a c=(-9)^{2}-4 \cdot 2 \cdot 7=81-56=25 $$
$$x_{1}=\frac{-b+\sqrt{D}}{2 a}=\frac{9+\sqrt{25}}{4}=\frac{9+5}{4} = \frac{14}{4} = \frac{7}{2} = 3.5; x_{2}=\frac{9-\sqrt{25}}{4} = \frac{9-5}{4} = \frac{4}{4} = 1.$$
	extbf{ OTBET: } $x \in \{3.5 ; 1\} .$
\end{enumerate}

Задание 2, стр 60

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Задание 2, стр 60

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