If $x^2+bx+c=0$ and $bx^2+cx+1=0$ share a root, then either $b + c + 1 = 0$ or $b^2 + c^2 + 1 =bc + b + c$

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.

Answer 1

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

Answer 2

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 $cy+1=b(y+1).$
Note that $b=c=1$ is also a possibility. Otherwise, $$-y=\dfrac{c-1}{b-1}=\dfrac{b-1}{b-c}$$ which simplifies exactly to what you need.