Question: If the quadratic equations $x^2+bx+c=0$ and $bx^2+cx+1=0$ have a common root then prove that either $b + c + 1 = 0$ or $b^2 + c^2 + 1 =bc + b + c$

Till yet,

I had figured the common root of the given two quadratic equation. i.e.

Multiplying first equation by $b$ and eliminating the term $bx^2$ from the equation I get the common root ($\alpha$ say),

$$\alpha=\frac{1 - cb}{b^2 - c}$$

Further putting this value in either of the equation didn't benefited me much.What it gave me was an odd, unfriendly equation. Can anyone help me in this?

Thanks in advance.


Let $y$ be the common root

So, we have $$y^2+by+c=0, by^2+cy+1=0$$

Solving for $y,y^2$ we get $y^2=\dfrac{b-c^2}{c-b^2},y=\dfrac{bc-1}{c-b^2}$

and using the identity $y^2=(y)^2$ we get $$b^3+c^3+1^3-3\cdot b\cdot c\cdot1=0$$

Now use Factorize the polynomial $x^3+y^3+z^3-3xyz$

  • $\begingroup$ Yep! Thank you very much. This was the very approach which I was seeking here, $$b^3+c^3+1-3bc=0$$ $$\implies (b+c+1)(b^2+c^2+1-bc-c-b)=0$$ now either $$b+c+1=0$$ or $$b^2+c^2+1=bc+b+c$$ $\endgroup$ – Saharsh Oct 6 '14 at 17:00

Let $y$ be the common root then, consider $y^2+by+c=0$ and $by^2+cy+1=0.$ Clearly $y\not=0.$ Therefore $$y^3+by^2+cy=0$$ $$y^3-1=0$$ $$(y-1)(y^2+y+1)=0.$$ If $y=1,$ then clearly $b+c+1=0.$
If $y^2+y+1=0,$ then note that $by+c=y+1$ and $cy+1=b(y+1).$ Solve this two simple linear equations and equate the value of $y.$

  • $\begingroup$ @Alpha: you are welcome. $\endgroup$ – Bumblebee Oct 7 '14 at 7:16

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