Quadratic equation roots

Question

How do I get the roots of the quadratic equation? $dx^2+(d ^2−d+1)x+d−1 = 0$ The solution is supposed to be $1-d$ and $-1/d$ but I have no idea how to get to it.

Answer 1

By the quadratic formula we get $$x_1=1-d$$ and $$x_2=-\frac{1}{d}$$ It is $$d_{1,2}=-\frac{d^2-d+1}{2d}\pm\sqrt{\left(\frac{d^2-d+1}{2d}\right)^2-\frac{d-1}{d}}$$

Answer 2

If $d=0$ we have $x=1$. Otherwise we follow Sridhara and multiply the equation by $4d$: $$4d^2x^2+4d(d^2-d+1)=4d(1-d)$$ and complete the square $$ 4d^2x^2+4d(d^2-d+1)+(d^2-d+1)^2=(d^2-d+1)^2+4d(1-d),$$ hence $$\begin{align}(2dx+d^2-d+1)^2&=(d^2-d)^2+2(d^2-d)+1-4(d^2-d)\\ &=(d^2-d)^2-2(d^2-d)+1\\ &=\bigl((d^2-1)-d\bigr)^2. \end{align}$$ Therefore $$2dx+d^2-d+1=\pm(d^2-1-d).$$ From here $x=-1/d$ and $x=1-d$ follows.

Answer 3

rearrange your equation like this $x^2+\frac{(d ^2−d+1)x}{d}+\frac{d−1}{d} = 0$

Let $\alpha$ and $\beta$ be roots of the equation

$\alpha + \beta = -d+1-\frac{1}{d}= 1-d-\frac{1}{d}=(-\frac{1}{d})+(1-d)$

and $\alpha\beta=\frac{1}{d}(d-1) \longleftrightarrow \alpha\beta=(-\frac{1}{d})(1-d)$

Now it is clearly visible what roots are $\alpha=(-\frac{1}{d})$ and $\beta=(1-d)$