A rectangle has length \( x + 5 \) and width \( x - 3 \). If the area is 120 square meters, find the positive value of \( x \), then compute the perimeter.
["Title: Solving for the Dimensions of a Rectangle: Given Area and Solving for ( x )", "In geometry, understanding how to work with rectangle dimensions when given area formulas is a key skill. In this article, we explore a practical problem: a rectangle with length ( x + 5 ) and width ( x - 3 ), where the area is 120 square meters. We’ll walk through finding the positive value of ( x ), then compute the perimeter.", "---", "### Problem Setup", "We are told:", "- Length = ( x + 5 )\n- Width = ( x - 3 )\n- Area = 120 square meters", "The area of a rectangle is calculated by multiplying length and width:", "[\n\ ext{Area} = (\ ext{Length}) \ imes (\ ext{Width})\n]", "Substitute the given expressions:", "[\n(x + 5)(x - 3) = 120\n]", "---", "### Step 1: Expand and Form a Quadratic Equation", "Expand the left-hand side using the distributive property (FOIL method):", "[\nx \cdot x + x \cdot (-3) + 5 \cdot x + 5 \cdot (-3) = x^2 - 3x + 5x - 15 = x^2 + 2x - 15\n]", "Now set equal to 120:", "[\nx^2 + 2x - 15 = 120\n]", "Move all terms to one side to form a standard quadratic equation:", "[\nx^2 + 2x - 15 - 120 = 0 \implies x^2 + 2x - 135 = 0\n]", "---", "### Step 2: Solve the Quadratic Equation", "Use the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Here, ( a = 1 ), ( b = 2 ), and ( c = -135 ). Plug in the values:", "[\nx = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-135)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 540}}{2} = \frac{-2 \pm \sqrt{544}}{2}\n]", "Simplify ( \sqrt{544} ):", "[\n\sqrt{544} = \sqrt{16 \cdot 34} = 4\sqrt{34}\n]", "So,", "[\nx = \frac{-2 \pm 4\sqrt{34}}{2} = -1 \pm 2\sqrt{34}\n]", "We are asked for the positive value of ( x ). Since ( \sqrt{34} \approx 5.83 ), then:", "[\nx = -1 + 2(5.83) \approx -1 + 11.66 = 10.66 > 0\n]", "Thus, the positive solution is:", "[\nx = -1 + 2\sqrt{34}\n]", "Though exact form is preferred, we can also approximate to verify:", "[\nx \approx 10.66\n]", "---", "### Step 3: Compute the Perimeter", "Perimeter ( P ) of a rectangle is:", "[\nP = 2(\ ext{Length} + \ ext{Width}) = 2[(x + 5) + (x - 3)] = 2(2x + 2) = 4x + 4\n]", "Now substitute ( x = -1 + 2\sqrt{34} ):", "[\nP = 4(-1 + 2\sqrt{34}) + 4 = -4 + 8\sqrt{34} + 4 = 8\sqrt{34}\n]", "Alternatively, using the approximate value:", "[\nP \approx 4(10.66) + 4 = 42.64 + 4 = 46.64 \ ext{ meters}\n]", "But the exact perimeter is:", "[\nP = 8\sqrt{34} \approx 46.64 \ ext{ meters}\n]", "---", "### Final Thoughts", "Solving for the unknown dimension ( x ) in a rectangle problem using area leads naturally to a quadratic equation. Factoring or the quadratic formula unlocks the solution, and from there, calculating perimeter is straightforward. This method is widely applicable in real-world geometry problems involving unknown side lengths.", "Remember:\n- Always verify that the width ( x - 3 > 0 ) (since width must be positive), and here (( 10.66 - 3 = 7.66 > 0 )) this holds.\n- Always simplify and confirm both exact and approximate answers when possible.", "---", "Summary:\n- Positive ( x ): ( x = -1 + 2\sqrt{34} )\n- Perimeter: ( 8\sqrt{34} ) meters", "This structured approach ensures accuracy and clarity in solving geometric problems involving variable dimensions and area."]
