Question

If the equation $x^2+bx+c=0$ and $b{x}^2+cx+1=0$ have a common root, then prove that either $b+c+1=0$ or $b^2+c^2+1=bc+b+c$

Answer 1

$x^2+bx+c=0$, $b{x}^2+cx+1=0\Rightarrow$ have one

common root, let it be ${}^{\prime }{\alpha }^{\prime }$.

${\alpha }^2+b\alpha +c=0;$ $b{\alpha }^2+c\alpha +1=0$

${\alpha }^2$ $\alpha$ $1$

b c $1$ b

c $1$ b c

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

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

${\left(\frac{bc-1}{c-b^2}\right)}^2=\frac{b-c^2}{c-b^2}$

$\frac{(bc-1{)}^2}{(c-b^2{)}^{\text{/}2}}=\frac{b-c^2}{\text{/}(c-b^2)}$

$b^2{\mathrm{c/}}^2+1-2bc=bc-b^3-c^3+b^2{\mathrm{c/}}^2$

$b^3+c^3+1-3bc\left(1\right)=0$

$(b+c+1)(b^2+c^2+1-bc-b-c)=0$

So, $b+c+0$

or

$b^2+c^2+1=b+c+bc$.