# Quadratic equation with weighted average coefficients

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?

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

$$\theta x^2-x+(1-\theta)=\theta (x^2-1)-(x-1)=(x-1)[\theta(x+1)-1]$$
$$θx^2-x+(1-θ)=0$$
Using, $$x = \frac{-b \pm\sqrt{b^2-4ac}}{2a}$$
$$x =\frac{-(-1) \pm\sqrt{(-1)^2-4\theta (1-\theta)}}{2\theta} = \frac{1 \pm\sqrt{1-4\theta +4\theta^2}}{2\theta} = \frac{1 \pm\sqrt{(1-2\theta)^2}}{2\theta} = \frac{1 \pm (1 - 2\theta) }{2} = (1- \theta) \ or \ \theta$$