Number of values of x

Question

$$a\dfrac{(x-b)(x-c)}{(a-b)(a-c)}+b\dfrac{(x-c)(x-a)}{(b-c)(b-a)}+c\dfrac{(x-a)(x-b)}{(c-a)(c-b)}=x$$

How many values of $x$ satisfy this equation? It is clear that x=a, x=b, x=c do satisfy the equation, but are those the only three possible solutions?

Answer 1

This has to be true for all $x$.

Firstly, there are at least $3$ solutions, assuming $a,b,c$ are distinct.

Secondly, this is a polynomial equation of degree $2$ in one variable $x$. In other words, you can rewrite the equation as $A x^2 + B x + C = 0$. But this can only have at most $2$ solutions, unless $A=B=C=0$.

Combining the two facts, the above equation holds for all $x$. Thus, if you multiply things out, it should come out as an identity.

Answer 2

If you move $x$ to the left side, you have a quadratic polynomial in $x$, that has the three zeroes $a,b,c$ that you have already found. However the only quadratic with more than two roots is the zero polynomial; hence this holds for all $x$.

Answer 3

Another way to see this: the expression on the left is the Lagrange polynomial of degree 2 passing through $(a,a)$, $(b,b)$, and $(c,c)$; in other words, it describes the line $y=x$. This value is always equal to $x$, which corresponds with the right-hand side.