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I have tried using the quadratic formula as well as factoring method to solve the following quadratic equation but failed to get the correct answer.

The equation is: $$ \theta x^2-x+(1-\theta)=0.$$

Note: the coefficient is theta

How to go ahead? What would be the two roots?

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  • $\begingroup$ $x=\frac{1\pm\sqrt{1-4\theta(1-\theta)}}{2}= \frac{1\pm (1-2\theta)}{2}$ $\endgroup$ – Qurultay May 4 at 9:44
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$\theta x^2-x+(1-\theta)=\theta (x^2-1)-(x-1)=(x-1)[\theta(x+1)-1]$

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    $\begingroup$ Makes sense, eventually, we get x=1 and x=1-θ/θ $\endgroup$ – user508281 May 4 at 9:51
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$θx^2-x+(1-θ)=0$

Using, $x = \frac{-b \pm\sqrt{b^2-4ac}}{2a}$

$x =\frac{-(-1) \pm\sqrt{(-1)^2-4\theta (1-\theta)}}{2\theta} = \frac{1 \pm\sqrt{1-4\theta +4\theta^2}}{2\theta} = \frac{1 \pm\sqrt{(1-2\theta)^2}}{2\theta} = \frac{1 \pm (1 - 2\theta) }{2} = (1- \theta) \ or \ \theta$

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