Quadratic Formula

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Quadratic Formula · Grade 9 · Worksheet 1

1. The quadratic function f(x) = 2x² - 8x + 3 can be written in vertex form as f(x) = a(x - h)² + k. What is the value of k? Answer: -5 Solution: To write the quadratic function f(x) = 2x² - 8x + 3 in vertex form f(x) = a(x - h)² + k, we need to complete the square. Factor out the coefficient of x² from the first two terms.
   Full step-by-step solution

   To write the quadratic function f(x) = 2x² - 8x + 3 in vertex form f(x) = a(x - h)² + k, we need to complete the square. Step 1: Factor out the coefficient of x² from the first two terms. f(x) = 2(x² - 4x) + 3 Step 2: Complete the square inside the parentheses. Take the coefficient of x inside the parentheses, which is -4. Divide it by 2: (-4)/2 = -2 Square the result: (-2)² = 4 Add and subtract this value inside the parentheses: f(x) = 2(x² - 4x + 4 - 4) + 3 Step 3: Group the perfect square trinomial and the constant. f(x) = 2[(x² - 4x + 4) - 4] + 3 f(x) = 2[(x - 2)² - 4] + 3 Step 4: Distribute the 2 and simplify. f(x) = 2(x - 2)² - 8 + 3 f(x) = 2(x - 2)² - 5 Now the function is in vertex form f(x) = a(x - h)² + k, where a = 2, h = 2, and k = -5. Therefore, the value of k is -5.

2. A local farmer is building a rectangular chicken coop against the side of her barn, so she only needs to use fencing for three sides. She has 40 meters of fencing material available. If the side perpendicular to the barn has a length of 'x' meters, write a quadratic function in standard form, A(x), that represents the area of the chicken coop in terms of x. Then, determine the maximum possible area she can enclose. Answer: A(x) = -2x² + 40x; Maximum area is 200 square meters Solution: We have a rectangular chicken coop against a barn, so only 3 sides need fencing: - Two sides perpendicular to the barn (each of length \( x \)) Total fencing used: \( x + x + L = 2x + L \) We have 40 m of fencing: \( 2x + L = 40 \) From \( 2x + L = 40 \), we get: \( L = 40 - 2x \) Area = length…
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We have a rectangular chicken coop against a barn, so only 3 sides need fencing: - Two sides perpendicular to the barn (each of length \( x \)) - One side parallel to the barn (length \( L \)) Total fencing used: \( x + x + L = 2x + L \) We have 40 m of fencing: \( 2x + L = 40 \) --- **Step 2: Express \( L \) in terms of \( x \)** From \( 2x + L = 40 \), we get: \( L = 40 - 2x \) --- **Step 3: Write the area \( A \) in terms of \( x \)** Area = length × width = \( L \times x \) Substitute \( L \): \( A(x) = (40 - 2x) \times x \) \( A(x) = 40x - 2x^2 \) This is already in standard form \( ax^2 + bx + c \): \( A(x) = -2x^2 + 40x \) --- **Step 4: Find the maximum possible area** Since \( a = -2 < 0 \), the parabola opens downward, so the vertex gives the maximum. For a quadratic \( ax^2 + bx + c \), the vertex occurs at: \( x = -\frac{b}{2a} \) Here \( a = -2 \), \( b = 40 \): \( x = -\frac{40}{2(-2)} = -\frac{40}{-4} = 10 \) So \( x = 10 \) m gives maximum area. --- **Step 5: Find \( L \) when \( x = 10 \)** \( L = 40 - 2x = 40 - 20 = 20 \) m --- **Step 6: Compute maximum area** \( A_{max} = L \times x = 20 \times 10 = 200 \) m² --- **Step 7: Final answer** The quadratic function is: \( A(x) = -2x^2 + 40x \) Maximum area is \( 200 \) square meters. --- **Answer:** A(x) = -2x² + 40x; Maximum area is 200 square meters

3. Aroha is helping her uncle, a structural engineer, design a parabolic arch for a new greenhouse. The shape of the arch can be modeled by the general quadratic equation ax² + bx + c = 0, where the solutions represent the x-intercepts (the points where the arch meets the ground). To find these critical points, Aroha needs to derive the quadratic formula from the standard form. Show all the steps to derive the quadratic formula x = [-b ± √(b²-4ac)] / 2a by completing the square on the general quadratic equation 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 to isolate the variable terms: ax² + bx = -c Divide every term by 'a' to make the coefficient of x² equal to 1: x² + (b/a)x = -c/a Complete the square.
   Full step-by-step solution

   Step 1: Start with the general quadratic equation: ax² + bx + c = 0 Step 2: Subtract 'c' from both sides to isolate the variable terms: ax² + bx = -c Step 3: Divide every term by 'a' to make the coefficient of x² equal to 1: x² + (b/a)x = -c/a Step 4: Complete the square. Take half of the coefficient of x, which is (b/a), so half is (b/2a). Square it to get (b²/4a²). Add this to both sides: x² + (b/a)x + (b²/4a²) = -c/a + (b²/4a²) Step 5: The left side is now a perfect square trinomial. Factor it: (x + b/2a)² = -c/a + b²/4a² Step 6: Combine the fractions on the right side. Find a common denominator, which is 4a²: (x + b/2a)² = (-4ac + b²) / 4a² Step 7: Rewrite the numerator: (x + b/2a)² = (b² - 4ac) / 4a² Step 8: Take the square root of both sides. Remember the ± sign: x + b/2a = ± √(b² - 4ac) / 2a Step 9: Subtract b/2a from both sides to isolate x: x = -b/2a ± √(b² - 4ac) / 2a Step 10: Combine the terms over the common denominator 2a: x = [-b ± √(b² - 4ac)] / 2a This is the quadratic formula.

4. 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 both sides by a: x² + (b/a)x = -c/a Complete the square. Take half of the coefficient of x: (b/a) ÷ 2 = b/(2a).
   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 both sides by a: x² + (b/a)x = -c/a Step 4: Complete the square. Take half of the coefficient of x: (b/a) ÷ 2 = b/(2a). 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: (x + b/(2a))² = -c/a + b²/(4a²) Step 6: Combine the right side over a common denominator: (x + b/(2a))² = (-4ac + b²)/(4a²) Step 7: Take the square root of both sides: x + b/(2a) = ±√(b² - 4ac)/(2a) Step 8: Isolate x by subtracting b/(2a) from both sides: x = -b/(2a) ± √(b² - 4ac)/(2a) Step 9: Combine the fractions: x = [-b ± √(b² - 4ac)] / (2a) The quadratic formula is derived as x = [-b ± √(b²-4ac)]/(2a).

5. Matiu is an engineer designing a parabolic reflector for a satellite dish. The cross-section of the reflector follows the path of a quadratic function. He knows that for any quadratic equation in the form ax² + bx + c = 0, the solutions can be found using a formula derived by completing the square. Starting with the general quadratic equation ax² + bx + c = 0, where a ≠ 0, derive the quadratic formula by completing the square. Show each step of the algebraic manipulation clearly. Answer: x = [-b ± sqrt(b² - 4ac)] / (2a) Solution: Start with the general quadratic equation in standard form. ax² + bx + c = 0 Subtract c from both sides to isolate the terms with x. ax² + bx = -c Divide every term by a (since a ≠ 0).
   Full step-by-step solution

   Step 1: Start with the general quadratic equation in standard form. ax² + bx + c = 0 Step 2: Subtract c from both sides to isolate the terms with x. ax² + bx = -c Step 3: Divide every term by a (since 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/a), square it, and add it to both sides. Half of b/a is b/(2a). Square it: (b/(2a))² = b²/(4a²) x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²) Step 5: The left side is now a perfect square trinomial. Factor it as (x + b/(2a))². (x + b/(2a))² = -c/a + b²/(4a²) Step 6: Combine the terms on the right side by finding a common denominator (4a²). -c/a = -4ac/(4a²) So the right side becomes: (-4ac + b²) / (4a²) = (b² - 4ac) / (4a²) (x + b/(2a))² = (b² - 4ac) / (4a²) Step 7: Take the square root of both sides. Remember to include both positive and negative roots. x + b/(2a) = ± sqrt((b² - 4ac) / (4a²)) Step 8: Simplify the square root on the right side. sqrt((b² - 4ac) / (4a²)) = sqrt(b² - 4ac) / sqrt(4a²) = sqrt(b² - 4ac) / (2a) x + b/(2a) = ± sqrt(b² - 4ac) / (2a) Step 9: Subtract b/(2a) from both sides to isolate x. x = -b/(2a) ± sqrt(b² - 4ac) / (2a) Step 10: Combine the terms on the right side since they have the same denominator (2a). x = [-b ± sqrt(b² - 4ac)] / (2a) The final answer is the quadratic formula: x = [-b ± sqrt(b² - 4ac)] / (2a)

6. Mere is designing a rectangular garden with a length that is 6 meters more than twice its width. The area of the garden is 140 square meters. Derive the quadratic formula from the standard form ax² + bx + c = 0 by completing the square to solve for the width of the garden. Answer: x = [-b ± sqrt(b² - 4ac)] / (2a) Solution: Start with the standard form of a quadratic equation: ax² + bx + c = 0, where a ≠ 0. Subtract c from both sides: ax² + bx = -c. Divide both sides by a (since a ≠ 0): x² + (b/a)x = -c/a.
   Full step-by-step solution

   Step 1: Start with the standard form of a quadratic equation: ax² + bx + c = 0, where a ≠ 0. Step 2: Subtract c from both sides: ax² + bx = -c. Step 3: Divide both sides by a (since a ≠ 0): 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 the left side as a perfect square: (x + b/(2a))² = -c/a + b²/(4a²). Step 6: Combine the right side over a common denominator 4a²: (x + b/(2a))² = (b² - 4ac)/(4a²). Step 7: Take the square root of both sides: x + b/(2a) = ± sqrt(b² - 4ac) / (2a). Step 8: Isolate x by subtracting b/(2a) from both sides: x = -b/(2a) ± sqrt(b² - 4ac)/(2a). Step 9: Combine the fractions: x = [-b ± sqrt(b² - 4ac)] / (2a). The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a).

7. A rectangular garden has a length that is 3 meters more than its width. If the area of the garden is 40 square meters, what is the width of the garden? Answer: 5 Solution: Let the width of the garden be \( w \) meters. The length is 3 meters more than the width, so length \( l = w + 3 \).
   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 3 meters more than the width, so length \( l = w + 3 \). --- **Step 2: Write the area equation** Area of a rectangle = length × width Given area = 40 square meters, so: \[ (w + 3) \times w = 40 \] --- **Step 3: Expand and rearrange** \[ w^2 + 3w = 40 \] Subtract 40 from both sides: \[ w^2 + 3w - 40 = 0 \] --- **Step 4: Solve the quadratic equation** We look for two numbers that multiply to -40 and add to 3. Those numbers are 8 and -5. So: \[ w^2 + 3w - 40 = (w + 8)(w - 5) = 0 \] --- **Step 5: Find possible values of w** From \( w + 8 = 0 \), we get \( w = -8 \) (not possible, width can't be negative). From \( w - 5 = 0 \), we get \( w = 5 \). --- **Step 6: Interpret the result** The width is 5 meters. Check: length = 5 + 3 = 8 meters, area = 8 × 5 = 40 square meters. Correct. --- **Final answer:** 5