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If quadratic equations $a_1x^2 + b_1x + c_1 = 0$ and $a_2x^2 + b_2x + c_2 = 0$ have both their roots common then they satisy, $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

But even if both the quadratic equations have only one common root say $\alpha$ then at $x=\alpha$
$$a_1\alpha^2 + b_1\alpha + c_1 = 0 \implies \frac{a_1}{c_1}\alpha^2 + \frac{b_1}{c_1}\alpha =-1$$ $$similarly$$ $$a_2\alpha^2 + b_2\alpha + c_2 = 0 \implies \frac{a_2}{c_2}\alpha^2 + \frac{b_2}{c_2}\alpha =-1$$

which on comparing gives me $$\frac{a_1}{c_1} = \frac{a_2}{c_2}, \space \frac{b_1}{c_1} = \frac{b_2}{c_2} \implies \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Where am I going wrong in understanding this?

marks_404
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    You can't equate coefficient with only one root $\alpha$. For example, you could have $\frac{a_1}{c_1}=\frac{a_2}{c_2}+1$, $\frac{b_1}{c_1}=\frac{b_2}{c_2}-\alpha$. – user10354138 Sep 07 '21 at 17:26
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    Your expression following "which on comparing gives me" is not justified. Assuming (as you have) that $0 \neq c_1, c_2$, in general the equation $K_1\alpha^2 + L_1\alpha = K_2\alpha^2 + L_2\alpha$ does not imply that $K_1 = K_2$ and $L_1 = L_2$. Remember, $\alpha$ is a fixed value. – user2661923 Sep 07 '21 at 17:33
  • What you get is a sufficient but not necessary condition. For the two pairs of ratios to be equal, you need the identity to hold for two distinct $\alpha$'s. –  Sep 07 '21 at 17:54
  • @IAmAGuest "What you get is a sufficient but not necessary condition" : did you intend "a necessary but not sufficient condition"? – user2661923 Sep 07 '21 at 18:08
  • @user2661923: I am talking about the OP's condition. –  Sep 07 '21 at 18:49
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    Try working with these equations which have only one common root. They might provide some insight.

    $$(x+1)(x-1)\quad =x^2-1\space\quad =x^2+0x-1 = 0\ (x-1)(x-1) \quad = (x-1)^2\quad = x^2+2x+1 = 0$$

     – poetasis Sep 10 '21 at 16:11

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