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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.

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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 $x^3+y^3+z^3-3xyz$

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  • $\begingroup$ 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
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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 $y.$

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

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