Quadratic by Factoring

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Quadratic by Factoring · Grade 9 · Worksheet 1

1. A physics class is designing a model rocket launch. The rocket's height above ground is modeled by the equation h(t) = -5t² + 40t + 20, where h is height in meters and t is time in seconds after launch. At what time will the rocket hit the ground? Answer: B. 8.47 seconds Solution: Set up the equation for when the rocket hits the ground: -5t² + 40t + 20 = 0 Divide all terms by -5 to simplify: t² - 8t - 4 = 0 Use the quadratic formula: t = [8 ± sqrt(64 + 16)] / 2 Calculate the discriminant: sqrt(80) = 4sqrt(5) ≈ 8.944 Calculate both solutions: t = (8 + 8.944)/2 ≈ 8.47 and t…
   Full step-by-step solution

   Step 1: Set up the equation for when the rocket hits the ground: -5t² + 40t + 20 = 0 Step 2: Divide all terms by -5 to simplify: t² - 8t - 4 = 0 Step 3: Use the quadratic formula: t = [8 ± sqrt(64 + 16)] / 2 Step 4: Calculate the discriminant: sqrt(80) = 4sqrt(5) ≈ 8.944 Step 5: Calculate both solutions: t = (8 + 8.944)/2 ≈ 8.47 and t = (8 - 8.944)/2 ≈ -0.47 Step 6: Discard the negative time value since time cannot be negative The correct answer is 8.47 seconds.

2. x² - 8x - 33 = 0 Answer: x = 11, x = -3 Solution: Factor the quadratic x² - 8x - 33 = 0. Find two numbers that multiply to -33 and add to -8. The numbers are -11 and 3 because (-11) × 3 = -33 and (-11) + 3 = -8.
   Full step-by-step solution

   Step 1: Factor the quadratic x² - 8x - 33 = 0. Step 2: Find two numbers that multiply to -33 and add to -8. The numbers are -11 and 3 because (-11) × 3 = -33 and (-11) + 3 = -8. Step 3: Write the factored form: (x - 11)(x + 3) = 0. Step 4: Apply the zero product property: x - 11 = 0 or x + 3 = 0. Step 5: Solve each equation: x = 11 or x = -3. The solutions are x = 11 and x = -3.

3. A right triangle is drawn on a coordinate plane with vertices at (0,0), (x,0), and (0,x+5). The area of this triangle is 42 square units. Write a quadratic equation in standard form that represents this situation, then solve for x by factoring. Answer: 7 Solution: The base of the triangle is the distance from (0,0) to (x,0), which is x units. The height of the triangle is the distance from (0,0) to (0,x+5), which is (x+5) units.
   Full step-by-step solution

   Step 1: The base of the triangle is the distance from (0,0) to (x,0), which is x units. Step 2: The height of the triangle is the distance from (0,0) to (0,x+5), which is (x+5) units. Step 3: The area of a right triangle is (1/2) * base * height. So, Area = (1/2) * x * (x+5). Step 4: Set the area equal to 42: (1/2) * x * (x+5) = 42. Step 5: Multiply both sides by 2 to eliminate the fraction: x(x+5) = 84. Step 6: Expand the left side: x^2 + 5x = 84. Step 7: Subtract 84 from both sides to set the equation to zero: x^2 + 5x - 84 = 0. Step 8: Factor the quadratic: We need two numbers that multiply to -84 and add to 5. These numbers are 12 and -7. Step 9: Write the factored form: (x + 12)(x - 7) = 0. Step 10: Set each factor equal to zero: x + 12 = 0 or x - 7 = 0. Step 11: Solve for x: x = -12 or x = 7. Step 12: Since a length cannot be negative in this geometric context, we discard x = -12. The answer is 7.

4. Tane is designing a rectangular mural for a community center. The area of the mural is 99 square feet. The length of the mural is 2 feet more than the width. What are the dimensions of the mural in feet? Answer: 9 and 11 Solution: Let w represent the width in feet. Then the length is w + 2 feet. Area = length × width, so w(w + 2) = 99.
   Full step-by-step solution

   Let w represent the width in feet. Then the length is w + 2 feet. Area = length × width, so w(w + 2) = 99. Expand: w^2 + 2w = 99. Subtract 99 from both sides: w^2 + 2w - 99 = 0. Factor: (w + 11)(w - 9) = 0. Apply zero product property: w + 11 = 0 or w - 9 = 0, so w = -11 or w = 9. Since width cannot be negative, w = 9. Then length = w + 2 = 11. The dimensions are 9 feet by 11 feet.

5. A rectangular garden has a length that is 5 meters longer than its width. If the area of the garden is 84 square meters, what are the dimensions of the garden? Answer: width = 7 meters, length = 12 meters Solution: Let the width of the garden be \( w \) meters. The length is 5 meters longer than the width, so length \( l = w + 5 \). Area of a rectangle = length × width.
   Full step-by-step solution

   Let's solve this step by step. --- **Step 1: Define variables** Let the width of the garden be \( w \) meters. The length is 5 meters longer 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. So: \[ w \times (w + 5) = 84 \] --- **Step 3: Expand and rearrange** \[ w^2 + 5w = 84 \] \[ w^2 + 5w - 84 = 0 \] --- **Step 4: Solve the quadratic equation** We solve \( w^2 + 5w - 84 = 0 \) by factoring. We look for two numbers whose product is -84 and whose sum is 5. The numbers 12 and -7 work because \( 12 \times (-7) = -84 \) and \( 12 + (-7) = 5 \). So: \[ w^2 + 5w - 84 = (w + 12)(w - 7) = 0 \] --- **Step 5: Find possible values of w** From \( (w + 12)(w - 7) = 0 \): - \( w + 12 = 0 \) → \( w = -12 \) (not valid, width can't be negative) - \( w - 7 = 0 \) → \( w = 7 \) So width \( w = 7 \) meters. --- **Step 6: Find length** Length \( l = w + 5 = 7 + 5 = 12 \) meters. --- **Step 7: Check** Area = \( 7 \times 12 = 84 \) square meters, which matches the problem. --- **Final answer:** Width = 7 meters, Length = 12 meters

6. A rectangular garden has an area of 54 square meters. The length of the garden is 3 meters more than twice its width. Liam needs to build a fence around the entire garden. What are the dimensions of the garden in meters? Answer: width = 4.5 m, length = 12 m Solution: width = w meters length = l meters - Area = 54 m² - Length is 3 more than twice the width.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Define variables** Let width = w meters length = l meters We are told: - Area = 54 m² - Length is 3 more than twice the width. So: l = 2w + 3 --- **Step 2: Write the area equation** Area = length × width 54 = l × w Substitute l from Step 1: 54 = (2w + 3) × w --- **Step 3: Expand and rearrange** 54 = 2w² + 3w Bring all terms to one side: 2w² + 3w - 54 = 0 --- **Step 4: Solve the quadratic equation** We can use the quadratic formula: w = [ -b ± sqrt(b² - 4ac) ] / (2a) Here a = 2, b = 3, c = -54. Discriminant: b² - 4ac = (3)² - 4(2)(-54) = 9 + 432 = 441 sqrt(441) = 21 So: w = [ -3 ± 21 ] / (4) --- **Step 5: Two possible solutions** First: w = ( -3 + 21 ) / 4 = 18 / 4 = 4.5 Second: w = ( -3 - 21 ) / 4 = -24 / 4 = -6 Width cannot be negative, so w = 4.5 m. --- **Step 6: Find length** l = 2w + 3 l = 2(4.5) + 3 l = 9 + 3 l = 12 m --- **Step 7: Check area** Area = 12 × 4.5 = 54 m² ✓ Length = 12, which is 3 more than twice 4.5 (since 2×4.5 = 9, plus 3 = 12) ✓ --- **Final answer:** width = 4.5 m, length = 12 m

7. A rectangular conference room has an area of 63 square meters. The length of the room is 2 meters less than twice its width. What are the dimensions of the room in meters? Answer: 7 and 9 Solution: When solving problems involving rectangular areas with given relationships between dimensions, we often use variables to represent the unknown measurements. The area formula for a rectangle is fundamental here.
   Full step-by-step solution

   When solving problems involving rectangular areas with given relationships between dimensions, we often use variables to represent the unknown measurements. The area formula for a rectangle is fundamental here. By expressing one dimension in terms of the other based on the given relationship, we can create a quadratic equation. Factoring this equation helps us find the possible dimensions that satisfy both the area requirement and the relationship between length and width.

8. A rectangular garden has an area of 54 square meters. The length of the garden is 3 meters more than twice its width. Find the width of the garden in meters. Answer: 4.5 Solution: Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: length \( l = 2w + 3 \).
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Define variables** Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: length \( l = 2w + 3 \). --- **Step 2: Write the area equation** Area of rectangle = length × width \( 54 = l \times w \) Substitute \( l = 2w + 3 \): \( 54 = (2w + 3) \times w \) --- **Step 3: Expand and rearrange** \( 54 = 2w^2 + 3w \) Bring all terms to one side: \( 2w^2 + 3w - 54 = 0 \) --- **Step 4: Solve the quadratic equation** Use the quadratic formula: \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Here \( a = 2 \), \( b = 3 \), \( c = -54 \). First, discriminant: \( D = b^2 - 4ac = 3^2 - 4(2)(-54) \) \( D = 9 + 432 = 441 \) \( \sqrt{D} = \sqrt{441} = 21 \) Now: \( w = \frac{-3 \pm 21}{2 \times 2} = \frac{-3 \pm 21}{4} \) --- **Step 5: Two possible solutions** Case 1: \( w = \frac{-3 + 21}{4} = \frac{18}{4} = 4.5 \) Case 2: \( w = \frac{-3 - 21}{4} = \frac{-24}{4} = -6 \) --- **Step 6: Choose the valid solution** Width cannot be negative, so \( w = 4.5 \) meters. --- **Step 7: Check** Length \( l = 2(4.5) + 3 = 9 + 3 = 12 \) Area = \( 12 \times 4.5 = 54 \) ✅ --- **Final answer:** Width = 4.5 meters