The Question:

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$...

Then find the roots of the equation $ax^2-bx(x-1)+c(x-1)^2=0$

My Attempt:

The new equation can be made into a quadratic as:


Now $$\text{Sum of roots}=\dfrac{-b}{a} = \dfrac{2c-b}{a-b+c}$$

And $$\text{Product of roots}=\dfrac{c}{a}=\dfrac{c}{a-b+c}$$

But I don't seem to be going anywhere with the way I'm proceeding

Please Help! Thanks!


3 Answers 3


We look at the equation $$ax^2-bx(x-1)+c(x-1)^2=0.\tag{1}$$ If $a\ne 0$, then $1$ is not a solution, so Equation (1) is equivalent to $$a\left(\frac{x}{x-1}\right)^2-b\frac{x}{x-1}+c=0.\tag{2}$$ If we put $y=-\frac{x}{x-1}$ then the solutions of (2) are the solutions of $ay^2+by+c=0$. It follows that $$\frac{-x}{x-1}=\alpha\quad\text{or}\quad \frac{-x}{x-1}=\beta.$$ Now we can easily solve for $x$.


hint: $\dfrac{2c-b}{a-b+c} = \dfrac{2\dfrac{c}{a}-\dfrac{b}{a}}{1-\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{2\alpha\cdot \beta+(\alpha+\beta)}{1+\alpha+\beta+\alpha\cdot \beta}$, can you continue?

  • $\begingroup$ Yeah...Thanks a lot! :) $\endgroup$
    – NeilRoy
    Jul 6, 2015 at 4:02

Let $\alpha = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$.

Note that $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$.

We want to calculate:

$$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} $$

In terms of $\alpha$ and $\beta$.

$$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} = \dfrac{2c-b\pm \sqrt{b^2-4bc+4c^2 -4ac + 4bc -4c^2}}{2(a-b+c)} = $$

$$= \dfrac{a}{a-b+c}\cdot\dfrac{2c-b\pm \sqrt{b^2-4ac}}{2a} =\dfrac{\frac{c}{a}}{1-\frac{b}{a}+\frac{c}{a}} +\dfrac{1}{1-\frac{b}{a}+\frac{c}{a}}\cdot\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} = $$

$$ = \dfrac{\alpha\beta}{1+\alpha+\beta + \alpha\beta} +\dfrac{1}{1+\alpha+\beta + \alpha\beta}\cdot\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$

So the roots are:

$$\dfrac{\alpha\beta + \alpha}{1+\alpha+\beta + \alpha\beta}\qquad \dfrac{\alpha\beta + \beta}{1+\alpha+\beta + \alpha\beta}$$


$$\dfrac{\alpha}{\alpha+1} \qquad \dfrac{\beta}{\beta+1}$$


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