#### Answer

$a=1, b=-2, \text { and } c=-1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the fact that a quadratic equation with roots $r_1$ and $r_2$ has a factored form of $(x-r_1)(x-r_2)=0$. Then use special products to convert the equation in the form $ax^2+bx+c=0.$
$\bf{\text{Solution Details:}}$
The factored form of the quadratic equation with the given roots $\{ 1+\sqrt{2},1-\sqrt{2} \},$ is \begin{array}{l}\require{cancel} (x-(1+\sqrt{2}))(x-(1-\sqrt{2}))=0 \\\\ (x-1-\sqrt{2})(x-1+\sqrt{2})=0 \\\\ [(x-1)-\sqrt{2}][(x-1)+\sqrt{2}]=0 .\end{array}
The expression above is a sum and difference of like terms with $(x-1)$ treated as the first term and $\sqrt{2}$ treated as the second term. Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent\begin{array}{l}\require{cancel} (x-1)^2-(\sqrt{2})^2=0 \\\\ (x-1)^2-2=0 .\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x^2-2x+1)-2=0 \\\\ x^2-2x+(1-2)=0 \\\\ x^2-2x-1=0 .\end{array}
Hence, $
a=1, b=-2, \text { and } c=-1
.$