Calculate x from a Square with Area 36: Solve for x > 0

When you solve $(x + 1)^2 = 36$, taking the square root gives $x + 1 = ±6$. This means x + 1 = 6 or x + 1 = -6, leading to x = 5 or x = -7.

Look at the given constraints! The problem states $x > 0$, so we must choose the positive value. Since x = -7 is negative, we reject it and keep x = 5.

If the area was something like 50, you'd get $x + 1 = ±\sqrt{50}$. You can simplify $\sqrt{50} = 5\sqrt{2}$ but still apply the same constraint x > 0.

No! Side lengths must always be positive in real-world geometry. That's why we have the constraint x > 0 - it ensures our side length x + 1 is positive.

Substitute x = 5 back into the problem: side length = 5 + 1 = 6, and area = $6^2 = 36$ ✓. Also check that x = 5 > 0 ✓.