Find x When (x+1)² Equals 36: Square Area Problem

First, let's recall the formula for calculating the area of a square with side length y (length units):

$A_{\boxed{}}=y^2$

Therefore, for a square with side length:

$x+1$

(length units), the expression for area is:

$$$A_{\boxed{}}=(x+1)^2$ (sq. units)

However the given data states that the square's area is 36 sq. units, meaning, in mathematical notation:

$A_{\boxed{}}=36$ (sq. units)

Therefore we can deduce the equation for the unknown x:

$(x+1)^2=36$

Let's continue and solve the resulting equation, starting by simplifying the expression on the left side,

To simplify the expression let's recall the formula for the square of a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

Let's continue and apply this formula to our equation, then combine like terms:

$(x+1)^2=36 \\ \downarrow\\ x^2+2\cdot x\cdot 1+1=36\\ x^2+2x-35=0$

We have obtained a quadratic equation, we can identify that the coefficient of the squared term is 1, therefore we can (try to) solve it using the quick factoring method,

We'll look for a pair of numbers whose product equals the constant term on the left side, and whose sum equals the coefficient of the first-degree term meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-35\\ m+n=2$

From the first requirement above, meaning the multiplication, we can conclude according to the rules of sign multiplication that the two numbers have opposite signs, and now we'll remember that the possible factor pairs of 35 are the numbers 7 and 5 or 35 and 1, fulfilling the second requirement mentioned, together with the fact that the numbers we're looking for have opposite signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=7\\ n=-5 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2+2x-35=0 \\ \downarrow\\ (x+7)(x-5)=0$

Where we used the pair of numbers we found earlier in this factoring,

We'll continue and consider the fact that on the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$x+7=0\\ \boxed{x=-7}$

Or:

$x-5=0\\ \boxed{x=5}$

However, from the domain of definition for x specified in the problem (which comes from the fact that the length of a side is positive):

x>-1 We can eliminate the solution: $x=-7$Therefore the only solution to the unknown in the problem that satisfies the given data is:

$\boxed{x=5}$

Therefore the correct answer is answer A.