Quadratic Formula

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Answer Key & Explanations

Quadratic Formula · Grade 9 · Worksheet 2

1. A rectangular community garden is being designed with a length that is 7 meters more than its width. The total area of the garden needs to be 170 square meters to accommodate all the planned vegetable beds. What is the width of the garden in meters? Answer: 10 Solution: Let the width be x meters. Since the length is 7 meters more than the width, the length is (x + 7) meters.
   Full step-by-step solution

   Step 1: Let the width be x meters. Since the length is 7 meters more than the width, the length is (x + 7) meters. Step 2: The area of a rectangle is length × width, so we have: x(x + 7) = 170 Step 3: Expand the equation: x² + 7x = 170 Step 4: Bring all terms to one side: x² + 7x - 170 = 0 Step 5: Factor the quadratic equation: (x + 17)(x - 10) = 0 Step 6: Solve for x: x = -17 or x = 10 Step 7: Since width cannot be negative, the width is 10 meters. The answer is 10.

2. A right triangle is drawn on a coordinate plane with vertices at (0,0), (x,0), and (0,8). The hypotenuse has length 17 units. Using the Pythagorean theorem, derive the quadratic equation that models this situation in standard form ax² + bx + c = 0. Answer: x² - 225 = 0 Solution: Step 1: Apply the Pythagorean theorem: (leg1)² + (leg2)² = (hypotenuse)² Step 2: Substitute the given values: x² + 8² = 17² Step 3: Calculate the squares: x² + 64 = 289 Step 4: Rearrange to standard form by subtracting 289 from both sides: x² + 64 - 289 = 0 Step 5: Simplify: x² - 225 = 0 Step 6:…
   Full step-by-step solution

   Step 1: Apply the Pythagorean theorem: (leg1)² + (leg2)² = (hypotenuse)² Step 2: Substitute the given values: x² + 8² = 17² Step 3: Calculate the squares: x² + 64 = 289 Step 4: Rearrange to standard form by subtracting 289 from both sides: x² + 64 - 289 = 0 Step 5: Simplify: x² - 225 = 0 Step 6: Identify coefficients: a = 1, b = 0, c = -225 The quadratic equation in standard form is x² - 225 = 0.

3. Derive x = [-b ± √(b²-4ac)] / (2a) by completing the square on ax² + bx + c = 0 Answer: x = [-b ± √(b²-4ac)] / (2a) Solution: Start with the general quadratic equation: ax² + bx + c = 0 Subtract c from both sides: ax² + bx = -c Divide through by a: x² + (b/a)x = -c/a Complete the square by adding (b/(2a))² to both sides: x² + (b/a)x + (b/(2a))² = -c/a + (b/(2a))² Write left side as perfect square: (x + b/(2a))² = -c/a…
   Full step-by-step solution

   Step 1: Start with the general quadratic equation: ax² + bx + c = 0 Step 2: Subtract c from both sides: ax² + bx = -c Step 3: Divide through by a: x² + (b/a)x = -c/a Step 4: Complete the square by adding (b/(2a))² to both sides: x² + (b/a)x + (b/(2a))² = -c/a + (b/(2a))² Step 5: Write left side as perfect square: (x + b/(2a))² = -c/a + b²/(4a²) Step 6: Combine right side terms: (x + b/(2a))² = (-4ac + b²)/(4a²) Step 7: Take square root of both sides: x + b/(2a) = ±√(b² - 4ac)/(2a) Step 8: Isolate x: x = [-b ± √(b² - 4ac)]/(2a) The quadratic formula is derived as x = [-b ± √(b²-4ac)]/(2a).

4. A construction company is designing a rectangular garden for a community center. The garden's length is 5 meters more than its width. If the area of the garden must be 84 square meters, what is the width of the garden in meters? Answer: 7 Solution: Let the width of the garden be \( w \) meters. The length is 5 meters more than the width, so length \( l = w + 5 \). Area of a rectangle = length × width.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Define the variables** Let the width of the garden be \( w \) meters. The length is 5 meters more than the width, so length \( l = w + 5 \). --- **Step 2: Write the area equation** Area of a rectangle = length × width. Given area = 84 square meters: \[ (w + 5) \times w = 84 \] --- **Step 3: Expand and rearrange** \[ w^2 + 5w = 84 \] Subtract 84 from both sides: \[ w^2 + 5w - 84 = 0 \] --- **Step 4: Solve the quadratic equation** We look for two numbers that multiply to -84 and add to 5. Possible factor pairs of 84: (1,84), (2,42), (3,28), (4,21), (6,14), (7,12). Since 12 and -7 multiply to -84 and add to 5? No, 12 + (-7) = 5 — yes, that works. So: \[ w^2 + 5w - 84 = (w + 12)(w - 7) = 0 \] --- **Step 5: Find possible values of w** \[ w + 12 = 0 \quad \text{or} \quad w - 7 = 0 \] \[ w = -12 \quad \text{or} \quad w = 7 \] --- **Step 6: Interpret the solution** Width cannot be negative, so \( w = -12 \) is not valid. Thus, \( w = 7 \) meters. --- **Step 7: Check** Width = 7 m, length = 7 + 5 = 12 m. Area = 7 × 12 = 84 m². Correct. --- **Final answer:** 7

5. A rectangular garden has a length that is 3 meters more than its width. If the area of the garden is 70 square meters, what is the width of the garden? Answer: 7 Solution: Let the width of the garden be \( w \) meters. The length is 3 meters more than the width, so length \( l = w + 3 \). Area of rectangle = length × width Given area = 70 square meters.
   Full step-by-step solution

   Let's solve step-by-step. --- **Step 1: Define the variables** Let the width of the garden be \( w \) meters. The length is 3 meters more than the width, so length \( l = w + 3 \). --- **Step 2: Write the area equation** Area of rectangle = length × width Given area = 70 square meters. So: \( (w + 3) \times w = 70 \) --- **Step 3: Expand and rearrange** \( w^2 + 3w = 70 \) \( w^2 + 3w - 70 = 0 \) --- **Step 4: Solve the quadratic equation** We solve \( w^2 + 3w - 70 = 0 \) by factoring. We need two numbers whose product is -70 and whose sum is 3. Possible pairs: (10, -7) → 10 × (-7) = -70, 10 + (-7) = 3. Yes. So: \( w^2 + 10w - 7w - 70 = 0 \) \( w(w + 10) - 7(w + 10) = 0 \) \( (w + 10)(w - 7) = 0 \) --- **Step 5: Find possible values of w** \( w + 10 = 0 \) → \( w = -10 \) (not possible, width can't be negative) \( w - 7 = 0 \) → \( w = 7 \) --- **Step 6: Verify** Width = 7 m, length = 7 + 3 = 10 m. Area = 10 × 7 = 70 m². Correct. --- **Final answer:** The width is 7 meters.

6. A drone is launched from a platform 12 meters high. Its height above ground is modeled by the function h(t) = -5t² + 20t + 12, where t is time in seconds. The drone needs to drop a package when it reaches its maximum height. How many seconds after launch should the package be dropped? Answer: 2 Solution: We are given the height function: h(t) = -5t² + 20t + 12. This is a quadratic function in the form h(t) = at² + bt + c, with a = -5, b = 20, c = 12.
   Full step-by-step solution

   We are given the height function: h(t) = -5t² + 20t + 12. This is a quadratic function in the form h(t) = at² + bt + c, with a = -5, b = 20, c = 12. Since a < 0, the parabola opens downward, so the maximum height occurs at the vertex. For a quadratic function at² + bt + c, the t-coordinate of the vertex is: t = -b / (2a) Substitute b = 20 and a = -5: t = -20 / (2 * -5) t = -20 / (-10) t = 20 / 10 t = 2 So the drone reaches maximum height at t = 2 seconds after launch. Therefore, the package should be dropped 2 seconds after launch.

7. Mere is a landscape architect designing a parabolic water fountain for a city park. The path of a water jet follows the quadratic equation y = 2x² - 16x + 30, where y is the height in meters and x is the horizontal distance from the nozzle. Mere needs to find the horizontal distances where the water jet is at ground level (y = 0). Instead of solving this specific equation directly, she wants to understand the general method for any quadratic. By completing the square on the general quadratic equation ax² + bx + c = 0 (where a ≠ 0), derive the quadratic formula x = [-b ± sqrt(b² - 4ac)] / (2a). Answer: x = [-b ± sqrt(b² - 4ac)] / (2a) Solution: Start with the general quadratic equation set equal to zero: ax² + bx + c = 0 Subtract c from both sides to isolate the x-terms: ax² + bx = -c Divide both sides by a (assuming a ≠ 0): x² + (b/a)x = -c/a Complete the square on the left side.
   Full step-by-step solution

   Step 1: Start with the general quadratic equation set equal to zero: ax² + bx + c = 0 Step 2: Subtract c from both sides to isolate the x-terms: ax² + bx = -c Step 3: Divide both sides by a (assuming a ≠ 0): x² + (b/a)x = -c/a Step 4: Complete the square on the left side. Take half of the coefficient of x, which is b/(2a), and square it: (b/(2a))² = b²/(4a²). Add this to both sides: x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²) Step 5: Write the left side as a perfect square trinomial: (x + b/(2a))² = -c/a + b²/(4a²) Step 6: Combine the terms on the right side over a common denominator 4a²: (x + b/(2a))² = (-4ac + b²) / (4a²) (x + b/(2a))² = (b² - 4ac) / (4a²) Step 7: Take the square root of both sides, remembering the ±: x + b/(2a) = ± sqrt(b² - 4ac) / (2a) Step 8: Subtract b/(2a) from both sides: x = -b/(2a) ± sqrt(b² - 4ac) / (2a) Step 9: Combine the terms over the common denominator 2a: x = [-b ± sqrt(b² - 4ac)] / (2a) This is the quadratic formula, which gives the solutions for x in any quadratic equation of the form ax² + bx + c = 0.