+0  
 
0
47
1
avatar

12x3+53x2−34x+5=0

Guest Jul 5, 2017

Best Answer 

 #1
avatar+18261 
+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
Sort: 

1+0 Answers

 #1
avatar+18261 
+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

16 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details