Using quadratic equations to solve problems, with detailed solutions and explanations.
Problems with Solutions
Problem 1: Right triangle perimeter and hypotenuse
A right triangle has a perimeter of 24 cm and a hypotenuse of 10 cm. Find the sides \(x\) and \(y\), \(x > y\), that make the right angle of the triangle.
Solution to Problem 1:
- Draw a triangle with the given information:
- The perimeter of the triangle is 24, hence: \[ x + y + 10 = 24 \]
- Use Pythagoras' theorem for a right triangle: \[ x^2 + y^2 = 10^2 \]
- Solve the perimeter equation for \(y\): \[ y = 14 - x \]
- Substitute \(y\) into the Pythagoras equation: \[ x^2 + (14 - x)^2 = 10^2 \]
- Expand and simplify: \[ x^2 + (196 - 28x + x^2) = 100 \implies 2x^2 - 28x + 96 = 0 \]
- Divide by 2: \[ x^2 - 14x + 48 = 0 \]
- Discriminant: \[ \Delta = b^2 - 4ac = (-14)^2 - 4(1)(48) = 196 - 192 = 4 \]
- Quadratic formula solutions: \[ x_1 = \frac{14 + 2}{2} = 8, \quad x_2 = \frac{14 - 2}{2} = 6 \]
- Corresponding \(y\) values: \[ y_1 = 14 - 8 = 6, \quad y_2 = 14 - 6 = 8 \]
- Since \(x > y\), the sides are: \[ x = 8 \text{ cm}, \quad y = 6 \text{ cm} \]
- Check the solution: \[ \text{Hypotenuse } h = \sqrt{x^2 + y^2} = \sqrt{64 + 36} = 10 \text{ cm} \] \[ \text{Perimeter} = x + y + h = 8 + 6 + 10 = 24 \text{ cm} \]
Problem 2: Consecutive numbers sum of squares
The sum of the squares of two consecutive real numbers is 61. Find the numbers.
Solution to Problem 2:
- Let the numbers be \(x\) and \(x+1\): \[ x^2 + (x+1)^2 = 61 \]
- Expand and simplify: \[ x^2 + x^2 + 2x + 1 = 61 \implies 2x^2 + 2x - 60 = 0 \]
- Divide by 2: \[ x^2 + x - 30 = 0 \]
- Discriminant: \[ \Delta = b^2 - 4ac = 1 + 120 = 121 \]
- Quadratic formula solutions: \[ x_1 = \frac{-1 + 11}{2} = 5, \quad x_2 = \frac{-1 - 11}{2} = -6 \]
- Corresponding numbers: \[ x_1 = 5, \ x_1 + 1 = 6 \] \[ x_2 = -6, \ x_2 + 1 = -5 \]
- Check: \[ 5^2 + 6^2 = 25 + 36 = 61, \quad (-6)^2 + (-5)^2 = 36 + 25 = 61 \]
Problem 3: Rectangular garden dimensions
A rectangular garden has a length that is 4 meters greater than its width. If its total area is 96 square meters, find its dimensions.
Solution to Problem 3:
- Let the width of the garden be \(x\) meters. Then the length is \(x + 4\) meters.
- Using the area formula (Area = length \(\times\) width): \[ x(x + 4) = 96 \]
- Expand and rewrite in standard quadratic form: \[ x^2 + 4x - 96 = 0 \]
- Factor the quadratic expression: \[ (x + 12)(x - 8) = 0 \]
- Solve for \(x\): \[ x = -12 \quad \text{or} \quad x = 8 \]
- Since physical dimensions must be positive, we reject \(x = -12\), so \(x = 8\).
- The width is \(8\) m and the length is \(8 + 4 = 12\) m.
- Check: \(8 \times 12 = 96\) m², which matches the given area.
Problem 4: Product of consecutive odd integers
The product of two positive consecutive odd integers is 143. Find the integers.
Solution to Problem 4:
- Let the first positive odd integer be \(x\). The next consecutive positive odd integer is \(x + 2\).
- Set up the equation based on their product: \[ x(x + 2) = 143 \]
- Expand and write in standard form: \[ x^2 + 2x - 143 = 0 \]
- Calculate the discriminant: \[ \Delta = b^2 - 4ac = 2^2 - 4(1)(-143) = 4 + 572 = 576 \]
- Apply the quadratic formula (\(\sqrt{576} = 24\)): \[ x = \frac{-2 \pm 24}{2} \] \[ x_1 = \frac{22}{2} = 11, \quad x_2 = \frac{-26}{2} = -13 \]
- Since the problem specifies positive integers, we choose \(x = 11\).
- The integers are \(11\) and \(11 + 2 = 13\).
- Check: \(11 \times 13 = 143\).
Problem 5 (Challenging): Boat motion against and with a current
A boat travels upstream for 34 km and then returns downstream for 34 km. The total time for the round trip is 6 hours. If the speed of the current is 5 km/h, find the speed of the boat in still water.
Solution to Problem 5:
- Let the speed of the boat in still water be \(x\) km/h, where \(x > 5\).
- The speed of the boat upstream (against the current) is \(x - 5\) km/h, and downstream (with the current) is \(x + 5\) km/h.
- Write the expression for the total time (\(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)): \[ \frac{34}{x - 5} + \frac{34}{x + 5} = 6 \]
- Divide both sides by 2 to simplify: \[ \frac{17}{x - 5} + \frac{17}{x + 5} = 3 \]
- Multiply both sides by the common denominator \((x - 5)(x + 5) = x^2 - 25\) to clear fractions: \[ 17(x + 5) + 17(x - 5) = 3(x^2 - 25) \]
- Expand both sides: \[ 17x + 85 + 17x - 85 = 3x^2 - 75 \] \[ 34x = 3x^2 - 75 \]
- Rearrange into standard quadratic form: \[ 3x^2 - 34x - 75 = 0 \]
- Calculate the discriminant: \[ \Delta = (-34)^2 - 4(3)(-75) = 1156 + 900 = 2056 \]
- Apply the quadratic formula: \[ x = \frac{-(-34) \pm \sqrt{2056}}{2(3)} = \frac{34 \pm 2\sqrt{514}}{6} = \frac{17 \pm \sqrt{514}}{3} \]
- Since speed must be greater than the current speed (\(x > 5\)), we take the positive root: \[ x = \frac{17 + \sqrt{514}}{3} \approx 13.22 \text{ km/h} \]
- Check the solution: $$\text{Upstream time} = \frac{34}{13.22 - 5} = \frac{34}{8.22} \approx 4.13 \text{ hours}$$ $$\text{Downstream time} = \frac{34}{13.22 + 5} = \frac{34}{18.22} \approx 1.87 \text{ hours}$$ $$\text{Total time} \approx 4.13 + 1.87 = 6.00 \text{ hours. Matches given criteria.}$$