# Difference between revisions of "2013 AMC 12B Problems/Problem 6"

The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11, so both problems redirect to this page.

## Problem

If $c$ is a constant such that $9x^2+10x+c$ is equal to the square of a binomial, then what is $c$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

## Solution 1

If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$. Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$. Notice this is a circle with radius $0$, which only contains one point. We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$. So, the only point is $(5, -3)$, so the sum is $5 + (-3) = 2 \implies \boxed{\textbf{(B)}}$. ~ asdf334

## Solution 2

If we move every term including $x$ or $y$ to the LHS, we get $$x^2 - 10x + y^2 + 6y = -34.$$ We can complete the square to find that this equation becomes $$(x - 5)^2 + (y + 3)^2 = 0.$$ Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$. Equality holds when the value inside the parhentheses is equal to $0$. We find that $$(x,y) = (5,-3)$$ and the sum we are looking for is $$5+(-3)=2 \implies \boxed{\textbf{(B)}}.$$ - Honestly