## Elementary Linear Algebra 7th Edition

The equality $ax^2+bx+c=0$ is satisfied for all $x$ when $a=b=c=0$. We have to show that there can be no other choice of $a, b, c$ and we will do that by showing that for three particular values of $x$ this choice is unique. Check the step by step guide for details.
+Let us make a system of equation by choosing three different $x$-es and and show that this system has a unique solution $a=b=c=0$: For $x=0$ we have $a\times 0^2 + b\times 0 + c=0\Rightarrow c=0$ For $x=1$ we have $a\times 1^2 + b\times 1 + c =0\Rightarrow a+b+c=0$ For $x=-1$ we have $a(-1)^2 + b(-1)+c=0\Rightarrow a-b+c=0$ So now we have the following system: \begin{align*} a+b+c=&0\\ a-b+c=&0\\ c=&0 \end{align*} The last equation already gives the solution for $c$: $$c=0$$ Putting this into the firs two we get \begin{align*} a+b=&0\\ a-b=&0. \end{align*} Adding these two together we obtain $$a+a+b-b=0\Rightarrow 2a=0$$ and this gives a solution for $a$: $$a=0$$ Putting this into the first equation we have $$0+b=0\Rightarrow b=0.$$ Now to return to the original problem, the equality $ax^2+bx+c=0$ is satisfied for all $x$ when $a=b=c=0$. We have shown that there can be no other choice of $a, b, c$ by showing that for three particular values of $x$ this choice is unique.