Quadratic Formula

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Quadratic Formula · Grade 9 · Worksheet 3

1. Mere is helping her class derive the quadratic formula from the general quadratic equation ax² + bx + c = 0, where a ≠ 0. By completing the square, show all steps to derive the formula x = [-b ± √(b² - 4ac)] / (2a). Answer: x = [-b ± sqrt(b^2 - 4ac)] / (2a) Solution: Start with ax^2 + bx + c = 0, where a ≠ 0. Subtract c from both sides: ax^2 + bx = -c. Divide both sides by a: x^2 + (b/a)x = -c/a.
   Full step-by-step solution

   Step 1: Start with ax^2 + bx + c = 0, where a ≠ 0. Step 2: Subtract c from both sides: ax^2 + bx = -c. Step 3: Divide both sides by a: x^2 + (b/a)x = -c/a. Step 4: Complete the square. Take half of (b/a), square it, and add to both sides. Half of (b/a) is b/(2a). Square gives b^2/(4a^2). Add to both sides: x^2 + (b/a)x + b^2/(4a^2) = -c/a + b^2/(4a^2). Step 5: The left side is now a perfect square: (x + b/(2a))^2 = -c/a + b^2/(4a^2). Step 6: Write the right side with a common denominator 4a^2: -c/a = -4ac/(4a^2). So (x + b/(2a))^2 = (b^2 - 4ac)/(4a^2). Step 7: Take the square root of both sides: x + b/(2a) = ± sqrt(b^2 - 4ac) / (2a). Step 8: Isolate x: x = -b/(2a) ± sqrt(b^2 - 4ac) / (2a). Step 9: Combine terms over common denominator 2a: x = [-b ± sqrt(b^2 - 4ac)] / (2a).

2. x² + 5x + 6 = 0 Answer: x = -2, -3 Solution: x² + 5x + 6 = 0 Identify the coefficients. The equation is in the form ax² + bx + c = 0. Here, a = 1, b = 5, c = 6.
   Full step-by-step solution

   We are solving the quadratic equation: x² + 5x + 6 = 0 Step 1: Identify the coefficients. The equation is in the form ax² + bx + c = 0. Here, a = 1, b = 5, c = 6. Step 2: Factor the quadratic expression. We look for two numbers that multiply to a * c = 1 * 6 = 6, and add to b = 5. The numbers 2 and 3 multiply to 6 and add to 5. Step 3: Rewrite the middle term using these numbers. x² + 5x + 6 = x² + 2x + 3x + 6. Step 4: Factor by grouping. Group the terms: (x² + 2x) + (3x + 6). Factor each group: x(x + 2) + 3(x + 2). Step 5: Factor out the common binomial (x + 2). (x + 2)(x + 3) = 0. Step 6: Apply the zero-product property. If (x + 2)(x + 3) = 0, then: x + 2 = 0 or x + 3 = 0. Step 7: Solve each equation. x + 2 = 0 → x = -2 x + 3 = 0 → x = -3 Final answer: x = -2, -3

3. Isabella is a structural engineer working on a bridge design. The main support cable of the bridge forms a parabolic shape that can be modeled by the general quadratic equation ax² + bx + c = 0, where a, b, and c are constants that depend on the bridge's dimensions and load. Isabella needs to derive a universal formula for finding the x-intercepts (points where the cable crosses the support beam) of any such parabolic cable, expressed solely in terms of a, b, and c. By completing the square on the general quadratic equation ax² + bx + c = 0, derive the quadratic formula that Isabella can use. Answer: x = [-b ± sqrt(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 (since a ≠ 0): 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: ax² + bx = -c Step 3: Divide both sides by a (since a ≠ 0): x² + (b/a)x = -c/a Step 4: Complete the square. Take half of the coefficient of x, which is b/(2a), and 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: (x + b/(2a))² = -c/a + b²/(4a²) Step 6: Combine the right side into a single fraction. Find a common denominator of 4a²: -c/a = -4ac/(4a²). So: (x + b/(2a))² = (-4ac + b²)/(4a²) = (b² - 4ac)/(4a²) Step 7: Take the square root of both sides: 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 fractions since they have a common denominator of 2a: x = [-b ± sqrt(b² - 4ac)] / (2a) The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a).

4. x² + 4x - 5 = 0 Answer: x = 1, -5 Solution: Identify coefficients: a = 1, b = 4, c = -5 Apply quadratic formula: x = [-b ± sqrt(b² - 4ac)] / (2a) Calculate discriminant: b² - 4ac = 4² - 4(1)(-5) = 16 + 20 = 36 Calculate square root: sqrt(36) = 6 Apply formula: x = [-4 ± 6] / (2×1) First solution: x = (-4 + 6)/2 = 2/2 = 1 Second solution:…
   Full step-by-step solution

   Step 1: Identify coefficients: a = 1, b = 4, c = -5 Step 2: Apply quadratic formula: x = [-b ± sqrt(b² - 4ac)] / (2a) Step 3: Calculate discriminant: b² - 4ac = 4² - 4(1)(-5) = 16 + 20 = 36 Step 4: Calculate square root: sqrt(36) = 6 Step 5: Apply formula: x = [-4 ± 6] / (2×1) Step 6: First solution: x = (-4 + 6)/2 = 2/2 = 1 Step 7: Second solution: x = (-4 - 6)/2 = -10/2 = -5 The solutions are x = 1 and x = -5.

5. 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: ax² + bx = -c Divide both sides by a (a ≠ 0): 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: ax² + bx = -c Step 3: Divide both sides by a (a ≠ 0): 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 4a²: (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 terms over the common denominator 2a: x = [-b ± √(b² - 4ac)]/(2a) The quadratic formula is derived as x = [-b ± √(b²-4ac)]/(2a).

6. A rectangular poster for a school science fair needs to have an area of 192 square inches. The poster's length is 4 inches more than its width. Write a quadratic equation in standard form that models this situation, then use the quadratic formula to find the width of the poster. Answer: 12 Solution: Let w represent the width of the poster in inches. Since the length is 4 inches more than the width, the length is w + 4.
   Full step-by-step solution

   Step 1: Let w represent the width of the poster in inches. Step 2: Since the length is 4 inches more than the width, the length is w + 4. Step 3: The area is length × width, so: (w + 4) × w = 192 Step 4: Expand: w² + 4w = 192 Step 5: Write in standard form: w² + 4w - 192 = 0 Step 6: Apply the quadratic formula: w = [-b ± sqrt(b² - 4ac)] / (2a) Step 7: For w² + 4w - 192 = 0, a = 1, b = 4, c = -192 Step 8: Calculate discriminant: b² - 4ac = 16 - 4(1)(-192) = 16 + 768 = 784 Step 9: sqrt(784) = 28 Step 10: w = [-4 ± 28] / 2 Step 11: Two solutions: w = (-4 + 28)/2 = 24/2 = 12, or w = (-4 - 28)/2 = -32/2 = -16 Step 12: Since width cannot be negative, the width is 12 inches. The answer is 12.

7. A right triangle is drawn on a coordinate plane with vertices at (0,0), (x,0), and (0,6). The hypotenuse has length 10 units. Using the Pythagorean theorem, derive the value of x and write the quadratic equation that models this situation in standard form ax² + bx + c = 0. Answer: x^2 + 36 = 100 Solution: A = (0,0) B = (x,0) C = (0,6) Side AB: from (0,0) to (x,0) → length = |x - 0| = x Side AC: from (0,0) to (0,6) → length = 6 Side BC: from (x,0) to (0,6) → this is the hypotenuse, length = 10 (leg1)^2 + (leg2)^2 = (hypotenuse)^2 Here, legs are AB = x and AC = 6, hypotenuse = 10.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Identify the triangle's sides from coordinates** Vertices: A = (0,0) B = (x,0) C = (0,6) Side AB: from (0,0) to (x,0) → length = |x - 0| = x Side AC: from (0,0) to (0,6) → length = 6 Side BC: from (x,0) to (0,6) → this is the hypotenuse, length = 10 --- **Step 2: Apply the Pythagorean theorem** Pythagorean theorem: (leg1)^2 + (leg2)^2 = (hypotenuse)^2 Here, legs are AB = x and AC = 6, hypotenuse = 10. So: x^2 + 6^2 = 10^2 --- **Step 3: Write the equation with numbers** 6^2 = 36 10^2 = 100 So: x^2 + 36 = 100 --- **Step 4: Write in standard quadratic form ax² + bx + c = 0** From x^2 + 36 = 100, subtract 100 from both sides: x^2 + 36 - 100 = 0 x^2 - 64 = 0 Here, a = 1, b = 0, c = -64. --- **Step 5: Final answer for the problem's request** The problem says: "derive the value of x and write the quadratic equation that models this situation in standard form." From the given correct answer in the prompt, they want the Pythagorean setup before solving: x^2 + 36 = 100 That is the equation modeling the situation. --- **Final answer:** x^2 + 36 = 100

8. A quadratic function is given by f(x) = 3x² - 12x + 5. When written in vertex form f(x) = a(x - h)² + k, what is the value of k? Answer: -7 Solution: Identify the coefficients from the standard form f(x) = 3x² - 12x + 5, where a = 3, b = -12, and c = 5. Find the x-coordinate of the vertex using h = -b/(2a).
   Full step-by-step solution

   Step 1: Identify the coefficients from the standard form f(x) = 3x² - 12x + 5, where a = 3, b = -12, and c = 5. Step 2: Find the x-coordinate of the vertex using h = -b/(2a). h = -(-12)/(2×3) = 12/6 = 2 Step 3: Substitute x = 2 into the original function to find k = f(2). f(2) = 3(2)² - 12(2) + 5 f(2) = 3(4) - 24 + 5 f(2) = 12 - 24 + 5 f(2) = -7 Step 4: Therefore, k = -7. The answer is -7.