Question

Rezolvarea in C (numere complexe) a ecuatiei de gradul al doilea cu coeficienti reali
Ecuatii Bipatrate
1) $3x^{2} = 108$
2) $(x-2)^{2} = -36$
3) $(x-3)^{2} + 16 = 0$
4) $4x^{2} + 81 = 0$

Answer 1

 

$\displaystyle\\1)\\3x^2=108\\\\x^2=\frac{108}{3}\\\\x^2=36\\x_{12}=\pm\sqrt{36}=\pm6\\x_1=6\\x_2=-6\\\text{------------}\\\\2)\\(x-2)^2=-36\\(x-2)_{12}=\pm\sqrt{-36}=\pm6i \\(x-2)_1 = 6i~~\Longrightarrow~~x_1=2+6i\\(x-2)_2 = -6i~~\Longrightarrow~~x_2=2-6i\\\text{------------}\\\\3)\\(x-3)^2+16=0\\(x-3)^2=-16\\(x-3)_{12}=\pm\sqrt{-16}=\pm4i \\(x-3)_1 = 4i~~\Longrightarrow~~x_1=3+4i\\(x-3)_2 = -4i~~\Longrightarrow~~x_2=3-4i\\\text{------------}$

$\displaystyle\\\\4)\\4x^2+81=0\\4x^2=-81\\\\x^2=\frac{-81}{4}\\\\x_{12}=\pm\sqrt{\frac{-81}{4}}=\pm\frac{9i}{2}\\\\x_1=\frac{9i}{2}\\\\x_1=-\frac{9i}{2}$

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