Rozwiąż równanie kwadratowe

[KASIA22-2006]

a) 2x^+3x-2=0
b) x^-6x+9=0
c) x^-5x+22=0
tez znak^ oznacza do kwadratu bo nie wiem jak postawić 2 u góry!!!Sory

[Sarrus]

a) 2x^+3x-2=0
b) x^-6x+9=0
c) x^-5x+22=0
tez znak^ oznacza do kwadratu bo nie wiem jak postawić 2 u góry!!!Sory

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\(\displaystyle{ problem \ drugi \ ) \ x^{2} \ - \ 6x \ + \ 9 \ = \ 0}\)

\(\displaystyle{ \begin{cases} \Delta \ = \ b^{2} \ - \ 4ac \\ gdzie : \\ a \ = \ +1 \\ b \ = \ -6 \\ c \ = \ +9 \end{cases}}\)

\(\displaystyle{ \Leftrightarrow \ \Delta \ = \ \left( -6 \right)^{2} \ - \ 4 \cdot 1 \cdot \left( 9\right) \ = \ 36 \ - \ 36 \ = \ 0}\)

\(\displaystyle{ \Delta \ = \ 0 \ \Leftrightarrow \ x_{0} \ : \ x_{0} \ \in Re \ \wedge \ x_{0} \ = \ \frac {-b}{2a} \ = \ \frac { - \left( -6 \right)}{2 \cdot 1} \ = \ \frac {6}{2} \ = \ +3}\)

\(\displaystyle{ \Leftrightarrow \ \left( x \ - \ 3 \right)^{2} \ = \ 0}\)

\(\displaystyle{ pierwiastkuje \ rownanie \ obustronnie \ :}\)

\(\displaystyle{ \Rightarrow \ \sqrt { \left( x \ - \ 3 \right)^{2} } \ = \ 0}\)

\(\displaystyle{ \Leftrightarrow \ \left| x \ - \ 3 \right| \ = \ 0}\)

\(\displaystyle{ badam \ miejsca \ zerowe \ funkcji \ : \ \psi (x) \ = \ x \ - \ 3 \ ; \ :}\)

\(\displaystyle{ \psi (x) \ = \ 0 \ \Leftrightarrow \ x \ - \ 3 \ = \ 0 \ \Leftrightarrow \ x \ = \ 3}\)
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\(\displaystyle{ problem \ pierwszy \ ) \ 2x^{2} \ +3x \ - \ 2 \ = \ 0}\)

\(\displaystyle{ \Delta \ = \ 9 \ + \ 16 \ = \ 25}\)

\(\displaystyle{ \Delta \ > \ 0 \ \Leftrightarrow \ \left( x_{1} \ ; \ x_{2} \right) \ : \ x_{1} \ , \ x_{2} \ \in \ Re \ \wedge \ x_{1} \ \neq \ x_{2} \ \wedge \ x_{1} \ = \ \frac {-b \ - \ \sqrt\Delta }{2a} \ \wedge \ x_{2} \ = \ \frac {-b \ + \ \sqrt\Delta }{2a}}\)

\(\displaystyle{ \Leftrightarrow \ x_{1} \ = \ \frac {-3 \ - \ 5 }{4} \ = \ \frac {-8}{4} \ = \ -2}\)

\(\displaystyle{ x_{2} \ = \ \frac {-3 \ + \ 5}{4} \ = \ \frac {2}{4} \ = \ \frac {1}{2}}\)

\(\displaystyle{ \Leftrightarrow \ \left( x \ + \ 2 \right) \left( x \ - \ \frac {1}{2} \right) \ = \ 0}\)

\(\displaystyle{ Iloczyn \ jest \ rowny \ zeru \ \Leftrightarrow \ jeden \ z \ jego \ czynnikow \ jest \ rowny \ zeru}\)

\(\displaystyle{ \Leftrightarrow \ x \ + \ 2 \ = \ 0 \ \vee \ x \ - \ \frac {1}{2} \ = \ 0}\)

\(\displaystyle{ \Leftrightarrow \ x \ = \ -2 \ \vee \ x \ = \ \frac {1}{2}}\)

\(\displaystyle{ \Rightarrow \ x \ : \ x \ = \{ \ -2 \ ; \ \frac {1}{2} \}}\)
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\(\displaystyle{ problem \ trzeci \ analogicznie \ jak \ pierwszy \ .}\)