# Quadratic Equations - Problems (1)

Using quadratic equations to solve problems; detailed solutions and explanations are included.

## 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:

• We start by drawing a triangle with the given information • The perimeter of the triangle is 24, hence
x + y + 10 = 24
• It is a right triangle, use Pythagoras theorem to obtain.
x2 + y2 = 102
• Solve the equation x + y + 10 = 24 for y.
y = 14 - x
• Substitute y in the equation x2 + y2 = 102 by the expression obtained above.
x2 + (14 - x)2 = 102
• Expand the square, group like terms and write the above equation with the right side equal to zero.
2x2 -28x + 96 = 0
• Multiply all terms in the above equation by 1/2.
x2 -14x + 48 = 0
• Find the discriminant of the above quadratic equation.
Discriminant Δ = b2 - 4*a*c = 196 - 192 = 4
• Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = [ -b + √Δ ] / (2 a) = [ 14 + 2 ] / 2 = 8
x2 = [ -b - √Δ ] / (2 a) = [ 14 - 2 ] / 2 = 6
• use the equation y = 14 - x to find the corresponding value of y.
y1 = 14 - 8 = 6
y2 = 14 - 6 = 8
• Taking into account the condition x > y, the sides that make the right angle of the triangle are: x = 8 cm and y = 6 cm.
Hypotenuse h = √ (x2 + y2)
= √ (82 cm2 + 62 cm2)
= √(64 cm2 + 36 cm2)
= 10 cm, it agrees with the given value.
Perimeter = y + x + hypotenuse
= 8 cm + 6 cm + 10 cm
= 24 cm, it agrees with the given value.

### 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, of the rectangle.

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 x and x+1 be the two consecutive numbers. The sum of the square of x and x + 1 is equal to 61.
x2 + (x + 1)2 = 61
• Expand (x + 1)2, group like terms and write the above equation with the right side equal to zero.
2x2 + 2x - 60 = 0
• Multiply all terms in the above equation by 1/2.
x2 + x - 30 = 0
• Find the discriminant of the above quadratic equation.
Discriminant Δ = b2 - 4*a*c = 1 + 120 = 121
• Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = [ - b + √Δ ] / 2*a = [ -1 + 11 ] / 2 = 5
x2 = [ - b - √Δ ] / 2*a = [ -1 - 11 ] / 2 = -6
• First solution to the problem
first number: x1 = 5
second number: x1 + 1 = 6
• Second solution to the problem
first number: x2 = -6
second number: x2 + 1 = -5
first solution sum of squares: 52 + 62
= 25 + 36 = 61
second solution sum of squares: (-6)2 + (-5)2
= 36 + 25 = 61
The two solutions to the problem agree with the given information in the problem.

### Matched Problem 2:

The sum of the squares of two consecutive even real numbers is 52. Find the numbers.

Solutions to Matched Problems

## Solutions to Matched Problems

Solution to Matched Problem 1:

• The perimeter of the rectangle is 60 m, hence
2 x + 2 y = 60
• The area of the rectangle is 200 m2, hence
x y = 200
• Solve the equation 2 x + 2 y = 60 for y.
y = 30 - x
• Substitute y in the equation x y = 200 by the expression for y obtained above.
x(30 - x) = 200
• Multiply, group like terms and write the above equation with the right hand side equal to zero.
- x2 +30 x - 200 = 0
• Find the discriminant of the above quadratic equation.
Discriminant D = b2 - 4 a c = 900 - 800 = 100
• Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = ( -b + √D ) / (2 a) = (-30 + 10 ) / (- 2) = 10 m
x2 = ( -b - √D ) / (2 a) = (-30 - 10 ) / (- 2) = 20 m
• use y = 30 - x found above to find the corresponding value of y.
y1 = 30 - 10 = 20 m
y2 = 30 - 20 = 10 m
• Taking into account the condition x > y, the length x = 20 m and the width y = 10 m.
As an exercise, check the perimeter and the area.

Solution to Matched Problem 2:

• Let x and x + 2 (the difference between two consecutive even numbers is 2) be the two consecutive even numbers. The sum of the square of x and x + 2 is equal to 52, hence
x2 + (x + 2)2 = 52
• Expand (x + 2)2, group like terms and write the above equation with the right side equal to zero.
2x2 + 4x - 48 = 0
• Multiply all terms in the above equation by 1/2 to obtain the following equivalent equation.
x2 + 2 x - 24 = 0
• Find the discriminant of the above quadratic equation.
Discriminant D = b2 - 4 a c = 4 + 90 = 100
• Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = ( - b + √D ) / (2 a) = ( - 2 + 10 ) / 2 = 4
x2 = ( - b - √D ) / (2 a) = ( - 2 - 10 ) / 2 = - 6
• First solution to the problem
first number: x1 = 4
second number: x1 + 2 = 6
• Second solution to the problem
first number: x2 = - 6
second number: x2 + 2 = - 4
As an exercise check that the square of the two numbers, for each solution, is 52.