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question

0
148
2 tysm

Dec 15, 2018

#1
+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
+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

PartialMathematician Dec 15, 2018
edited by PartialMathematician  Dec 15, 2018
edited by PartialMathematician  Dec 15, 2018
#2
0

ywsm (you welcome so much)

PartialMathematician  Dec 15, 2018