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12x3+53x2−34x+5=0

Guest Jul 5, 2017

Best Answer 

 #1
avatar+19808 
+2

12x^3+53x^2−34x+5=0

 

\(\begin{array}{|l|rcll|} \hline & 12x^3+53x^2-34x+5 &=& 0 \\ \mathbf{ x_1 = -5} & 12\cdot (-5)^3 + 53\cdot (-5)^2 -34\cdot (-5) + 5 &=& 0 \\ & -1500+1325+170+5 &=& 0 \\ & 0 &=& 0\ \checkmark \\ \hline \end{array}\)

 

 

\(\begin{array}{|rcll|} \hline 12x^2-7x+1 &=& 0 \\ x &=& \frac{7\pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot 1 } }{2\cdot 12} \\ &=& \frac{7\pm \sqrt{1} }{24} \\ &=& \frac{7\pm 1 }{24} \\\\ x_2 &=& \frac{7 + 1 }{24} \\ \mathbf{ x_2 } & \mathbf{=}& \mathbf{ \frac{1 }{3} } \\\\ x_3 &=& \frac{7 - 1 }{24} \\ \mathbf{ x_3 } & \mathbf{=}& \mathbf{\frac{1 }{4}} \\ \hline \end{array}\)

 

\(x_1 = -5 \\ x_2 = \frac{1 }{3} \\ x_3 = \frac{1 }{4} \)

 

laugh

heureka  Jul 5, 2017
 #1
avatar+19808 
+2
Best Answer

12x^3+53x^2−34x+5=0

 

\(\begin{array}{|l|rcll|} \hline & 12x^3+53x^2-34x+5 &=& 0 \\ \mathbf{ x_1 = -5} & 12\cdot (-5)^3 + 53\cdot (-5)^2 -34\cdot (-5) + 5 &=& 0 \\ & -1500+1325+170+5 &=& 0 \\ & 0 &=& 0\ \checkmark \\ \hline \end{array}\)

 

 

\(\begin{array}{|rcll|} \hline 12x^2-7x+1 &=& 0 \\ x &=& \frac{7\pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot 1 } }{2\cdot 12} \\ &=& \frac{7\pm \sqrt{1} }{24} \\ &=& \frac{7\pm 1 }{24} \\\\ x_2 &=& \frac{7 + 1 }{24} \\ \mathbf{ x_2 } & \mathbf{=}& \mathbf{ \frac{1 }{3} } \\\\ x_3 &=& \frac{7 - 1 }{24} \\ \mathbf{ x_3 } & \mathbf{=}& \mathbf{\frac{1 }{4}} \\ \hline \end{array}\)

 

\(x_1 = -5 \\ x_2 = \frac{1 }{3} \\ x_3 = \frac{1 }{4} \)

 

laugh

heureka  Jul 5, 2017

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