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1) 3x^2 + 81=0

2) 6x^2 = -126

3) (1/4)x^2 + 12=0

 

Solve by using this symbol "±​".

 Aug 15, 2018
edited by Mathrules  Aug 15, 2018
edited by asinus  Aug 15, 2018
 #1
avatar+21848 
+2

1)

3x^2 + 81=0

 

\(\begin{array}{|rcll|} \hline 3x^2 + 81 &=& 0 \quad | \quad : 3 \\ x^2 + 27 &=& 0 \quad | \quad -27 \\ x^2 &=& -27 \\ x^2 &=& (-1)\cdot 9 \cdot 3 \\ x &=& \pm\sqrt{(-1)\cdot 9 \cdot 3} \\ x &=&\pm\sqrt{9}\sqrt{3}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm 3\sqrt{3}\cdot i \\ \hline \end{array}\)

 

laugh

 Aug 15, 2018
edited by heureka  Aug 15, 2018
 #2
avatar+21848 
+2

2)
6x^2 = -126

 

\(\begin{array}{|rcll|} \hline 6x^2 &=& -126 \quad | \quad : 6 \\ x^2 &=& -21 \\ x^2 &=& (-1)\cdot 21 \\ x &=& \pm\sqrt{(-1)\cdot 21} \\ x &=&\pm \sqrt{21}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm \sqrt{21}\cdot i \\ \hline \end{array}\)

 

laugh

 Aug 15, 2018
edited by heureka  Aug 15, 2018
 #3
avatar+21848 
+2

3)
(1/4)x^2 + 12=0

 

\(\begin{array}{|rcll|} \hline (1/4)x^2 + 12 &=& 0 \quad | \quad \cdot 4 \\ x^2 + 48 &=& 0 \quad | \quad -48 \\ x^2 &=& -48 \\ x^2 &=& (-1)\cdot 16 \cdot 3 \\ x &=& \pm \sqrt{(-1)\cdot 16 \cdot 3} \\ x &=&\pm\sqrt{16}\sqrt{3}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm 4\sqrt{3}\cdot i \\ \hline \end{array}\)

 

laugh

 Aug 15, 2018

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