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tysm

 Dec 15, 2018

Best Answer 

 #1
avatar+701 
+1

\(6x^2 = 2x-1\).

 

Subtract both sides by 2x - 1 to get 6x^2 - 2x + 1 = 0. Using the quadratic formula

(\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)) with a = 6, b = -2, and c = 1, we have \(x = \dfrac{-(-2)\pm \sqrt{(-2)^2-4(6)(1)}}{2(6)}\), which simplifies into \(x = \dfrac{2\pm \sqrt{4 - 24}}{12} \Rightarrow x = \dfrac{2\pm \sqrt{-20}}{12}\). Our answer is \(x = \dfrac{2\pm \sqrt{-20}}{12}\), which is the same as \(\boxed{ x = \dfrac{1\pm i\sqrt{5}}{6}}\) . Note that \(i = \sqrt{-1}\).

 

Hope this helps, 

 

- PM

 Dec 15, 2018
edited by PartialMathematician  Dec 15, 2018
edited by PartialMathematician  Dec 15, 2018
 #1
avatar+701 
+1
Best Answer

\(6x^2 = 2x-1\).

 

Subtract both sides by 2x - 1 to get 6x^2 - 2x + 1 = 0. Using the quadratic formula

(\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)) with a = 6, b = -2, and c = 1, we have \(x = \dfrac{-(-2)\pm \sqrt{(-2)^2-4(6)(1)}}{2(6)}\), which simplifies into \(x = \dfrac{2\pm \sqrt{4 - 24}}{12} \Rightarrow x = \dfrac{2\pm \sqrt{-20}}{12}\). Our answer is \(x = \dfrac{2\pm \sqrt{-20}}{12}\), which is the same as \(\boxed{ x = \dfrac{1\pm i\sqrt{5}}{6}}\) . Note that \(i = \sqrt{-1}\).

 

Hope this helps, 

 

- PM

PartialMathematician Dec 15, 2018
edited by PartialMathematician  Dec 15, 2018
edited by PartialMathematician  Dec 15, 2018
 #2
avatar+701 
0

ywsm (you welcome so much)

PartialMathematician  Dec 15, 2018

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