Expanding $(x+m)(x+n)$ gives $x^2+(m+n)x+mn$. Since $m+n=11$, the coefficient of $x$ must be $11$. That eliminates any expression without middle term $11x$. Also, because $m$ and $n$ are positive integers that add to $11$, one possible pair is $5$ and $6$. Their product is $30$, so the quadratic could be $x^2+11x+30$.

A parabola in vertex form is $y=a(x-h{)}^2+k$, where the vertex is $(h,k)$. Since the vertex is $(2,-3)$, the equation must be $y=a(x-2{)}^2-3$. Because the parabola opens upward, $a$ must be positive. The equation that matches is $y=(x-2{)}^2-3$.

Since both expressions equal $y$, set them equal to each other: $x^2=4x$. Then solve:

So the solutions are $x=0$ and $x=4$. The question asks for the solution where $x\ne 0$, so the value of $x$ is $4$.

Factor the equation:

So the solutions are $3$ and $-3$. Their sum is $3+(-3)=0$, so this statement must be true.

For a square, area equals side length squared. Since $A(x)=x^2+6x+9=(x+3{)}^2$, the side length is $x+3$.