Two quadratic equations having a common root

Question

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 satisfy $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$

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

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?

Answer 1

You cannot always simply “compare” the corresponding coefficients of $2$ polynomial equations of the same degree when they have common roots. It can only be done when both the equations are exactly the same, i.e., the given polynomials are scalar multiples of each other.

That is, comparing coefficients is valid when both the polynomial equations have all roots(no. of common roots is the same as the degree of the polynomial) common, since in this case, both the equations are exactly the same by the factor theorem.

For example, let’s consider deriving the condition for $2$ quadratic equations $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ to have both roots common, say $\alpha$ and $\beta$. Since a quadratic equation has $2$ roots, these equations can be rewritten in factor form as

$$a_1(x-\alpha)(x-\beta)=0$$ $$a_2(x-\alpha)(x-\beta)=0$$

This means that both these equations are the same, so both the quadratic polynomials $a_1x^2+b_1x+c_1$ and $a_2x^2+b_2x+c_2$ only differ by a non-zero scalar, i.e., the coefficients of $x^2$, $x$ and $x^0$ are in the same ratio. This is why we can compare the coefficients in this case:

$$\frac{a_1}{a_2}= \frac{b_1}{b_2} = \frac{c_1}{c_2} $$

But when both equations have only $1$ root common, say $\alpha$, they would be of the form:

$$a_1(x-\alpha)(x-\beta)=0$$ $$a_2(x-\alpha)(x-\gamma)=0$$

Since $\beta\ne\gamma$ as already stated, both these equations are not the same, hence we cannot compare their coefficients.

So, to derive the condition for $2$ quadratic equations to have only one common root, consider putting $x=\alpha$ in both the equations:

$$a_1\alpha^2+b_1\alpha+c_1=0$$ $$a_2\alpha^2+b_2\alpha+c_2=0$$

This can be recognized as a system of linear equations in $2$ variables. By the cross-multiplication method/Cramer’s rule:

$$\frac{\alpha^2}{b_1c_2-b_2c_1}=\frac{\alpha}{c_1a_2-c_2a_1}=\frac1{a_1b_2-a_2b_1}$$

Eliminating $\alpha$, we get:

$$\left(\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\right)^2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}$$ $$\therefore\boxed{(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)}$$

Answer 2

As you state, it is a proven fact that if two quadratic equations have the same two roots, each pair of corresponding coefficients (the two $x^2$ coefficients, the two coefficients of $x$, and the two constant terms) have the same ratio.

The key step in your argument is that from

$$\frac{a_1}{c_1}\alpha^2+\frac{b_1}{c_1}\alpha=-1 \quad\text{and}\quad \frac{a_2}{c_2}\alpha^2+\frac{b_2}{c_2}\alpha=-1 \tag1$$

you infer that $\dfrac{a_1}{c_1} = \dfrac{a_2}{c_2}$ and $\dfrac{b_1}{c_1} = \dfrac{b_2}{c_2}.$ But note that the two equations on line $(1)$ are really just quadratic equations in the variable $\alpha,$ known to be equal at one particular value of $\alpha$ but not necessarily at others, and you're saying that since the ratio of the constant terms in these two equations is $1$ the ratio of the other coefficients must be $1$ as well.

We know that if we have two equal roots then these three ratios will all be the same, but you're saying you can do this with just one equal root. You're essentially assuming the surprising "fact" that you found in order to find that "fact." This is an example of circular reasoning.

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But often the best way to disprove a false "theorem" is with a counterexample.

As suggested in a comment, try the two quadratic equations with coefficients $a_1 = 1,$ $b_1 = 0,$ $c_1 = -1$ and $a_2 = 1,$ $b_2 = -2,$ $c_2 = 1.$ These equations are \begin{align} x^2 \phantom{{} + 2x} - 1 &= 0,\\ x^2 - 2x + 1 &= 0. \end{align}

Both equations have a root at $x = 1.$ So let's work out the equations on line $(1)$ with $\alpha = 1.$

\begin{align} \frac{a_1}{c_1}\alpha^2+\frac{b_1}{c_1}\alpha &= \frac{1}{-1}1^2+\frac{0}{-1}1 = -1 + 0 = -1, \\ \frac{a_2}{c_2}\alpha^2+\frac{b_2}{c_2}\alpha &= \frac{1}{1}1^2+\frac{-2}{1}1 = 1 - 2 = -1. \end{align}

So both equations are satisfied, but in different ways: $\dfrac{a_1}{c_1}\neq \dfrac{a_2}{c_2}$ and $\dfrac{b_1}{c_1}\neq \dfrac{b_2}{c_2}.$