Square of sum: Using short multiplication formulas

To find $x$ in the given triangle, let's apply the Pythagorean Theorem. The squared lengths of the triangle's legs and hypotenuse are related by this equation:

$(x+1)^2 + x^2 = (x+2)^2$

First, expand each term:

• $(x+1)^2 = x^2 + 2x + 1$
• $x^2 = x^2$
• $(x+2)^2 = x^2 + 4x + 4$

Plug these into the Pythagorean Theorem equation:

$(x^2 + 2x + 1) + x^2 = x^2 + 4x + 4$

Combine like terms:

$2x^2 + 2x + 1 = x^2 + 4x + 4$

Rearrange the equation to isolate terms on one side:

$2x^2 + 2x + 1 - x^2 - 4x - 4 = 0$

Simplify to get a quadratic equation:

$x^2 - 2x - 3 = 0$

Now, solve for $x$ using factoring. Look for two numbers that multiply to $-3$ and add to $-2$. These numbers are $-3$ and $1$:

$(x - 3)(x + 1) = 0$

Set each factor equal to zero:

• $x - 3 = 0 \Rightarrow x = 3$
• $x + 1 = 0 \Rightarrow x = -1$

Given the condition $x > 0$, the valid solution is:

$x = 3$

To find the area of a square with side length $x + 1$, we apply the formula for the area of a square, which is side squared. This means we need to calculate $(x + 1)^2$.

Here are the steps to solve the problem:

• Step 1: Identify the expression for the side length. The side length of the square is given as $x + 1$.
• Step 2: Use the formula for the area of a square: $(\text{side})^2$.
• Step 3: Substitute the side length with $x + 1$: $(x + 1)^2$.
• Step 4: Expand the expression using the formula for the square of a sum: $(a + b)^2 = a^2 + 2ab + b^2$, where $a = x$ and $b = 1$.
• Step 5: Calculation:
• $(x + 1)^2 = x^2 + 2(x)(1) + 1^2$
• $(x + 1)^2 = x^2 + 2x + 1$

Therefore, the algebraic expression for the area of the square is $x^2 + 2x + 1$.

To solve this problem, follow these steps:

• Step 1: Set up the equation based on the given information. The side of the square is given by $x + 1$, and the area formula for a square is $(\text{side})^2$. Thus, we set up the equation as $(x + 1)^2 = 36$.
• Step 2: Solve for $x$. Take the square root of both sides of the equation:

$(x + 1) = \pm \sqrt{36}$
$(x + 1) = \pm 6$

• Since $x > 0$ is a condition, only the positive value of $x + 1$ is valid, so $x + 1 = 6$.
• Solve for $x$ by subtracting 1 from both sides:

$x = 6 - 1$
$x = 5$

Therefore, the solution to the problem is $x = 5$.