Quadratic Equations - Problems (1)
Using quadratic equations to solve problems, with detailed solutions and explanations.
Problems with Solutions
Problem 1:
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:
\[
\text{Hypotenuse } h = \sqrt{x^2 + y^2} = \sqrt{64 + 36} = 10 \text{ cm}
\]
\[
\text{Perimeter} = x + y + h = 8 + 6 + 10 = 24 \text{ cm}
\]
Matched Problem 1:
A rectangle has a perimeter of 60 m and an area of 200 m². Find the length \(x\) and width \(y\), \(x > y\).
Solutions to Matched Problems
Problem 2:
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:
\[
5^2 + 6^2 = 25 + 36 = 61, \quad (-6)^2 + (-5)^2 = 36 + 25 = 61
\]
Matched Problem 2:
The sum of the squares of two consecutive even numbers is 52. Find the numbers.
Solutions to Matched Problems
Solutions to Matched Problems
Solution to Matched Problem 1:
- Perimeter equation:
\[
2x + 2y = 60 \implies y = 30 - x
\]
- Area equation:
\[
xy = 200 \implies x(30-x) = 200 \implies -x^2 + 30x - 200 = 0
\]
- Discriminant:
\[
D = b^2 - 4ac = 900 - 800 = 100
\]
- Quadratic solutions:
\[
x_1 = \frac{-30 + 10}{-2} = 10, \quad x_2 = \frac{-30 - 10}{-2} = 20
\]
- Corresponding \(y\) values:
\[
y_1 = 30 - 10 = 20, \quad y_2 = 30 - 20 = 10
\]
- With \(x > y\):
\[
x = 20 \text{ m}, \quad y = 10 \text{ m}
\]
Solution to Matched Problem 2:
- Let the numbers be \(x\) and \(x+2\):
\[
x^2 + (x+2)^2 = 52
\]
- Expand and simplify:
\[
x^2 + x^2 + 4x + 4 = 52 \implies 2x^2 + 4x - 48 = 0 \implies x^2 + 2x - 24 = 0
\]
- Discriminant:
\[
D = b^2 - 4ac = 4 + 96 = 100
\]
- Quadratic formula solutions:
\[
x_1 = \frac{-2 + 10}{2} = 4, \quad x_2 = \frac{-2 - 10}{2} = -6
\]
- Corresponding numbers:
\[
x_1 = 4, \ x_1 + 2 = 6
\]
\[
x_2 = -6, \ x_2 + 2 = -4
\]
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