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.
x^{2} + y^{2} = 10^{2}

Solve the equation x + y + 10 = 24 for y.
y = 14 - x

Substitute y in the equation x^{2} + y^{2} = 10^{2} by the expression obtained above.
x^{2} + (14 - x)^{2} = 10^{2}

Expand the square, group like terms and write the above equation with the right side equal to zero.
2x^{2} -28x + 96 = 0

Multiply all terms in the above equation by 1/2.
x^{2} -14x + 48 = 0

Find the discriminant of the above quadratic equation.
Discriminant Δ = b^{2} - 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.
Check answer:
Hypotenuse h = √ (x^{2} + y^{2})
= √ (8^{2} cm^{2} + 6^{2} cm^{2})
= √(64 cm^{2} + 36 cm^{2})
= 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 m^{2}. Find the length x and width y, x > y, of the rectangle.

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.
x^{2} + (x + 1)^{2} = 61

Expand (x + 1)^{2}, group like terms and write the above equation with the right side equal to zero.
2x^{2} + 2x - 60 = 0

Multiply all terms in the above equation by 1/2.
x^{2} + x - 30 = 0

Find the discriminant of the above quadratic equation.
Discriminant Δ = b^{2} - 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
Check answer:
first solution sum of squares: 5^{2} + 6^{2}
= 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.

The perimeter of the rectangle is 60 m, hence
2 x + 2 y = 60

The area of the rectangle is 200 m^{2}, 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.
- x^{2} +30 x - 200 = 0

Find the discriminant of the above quadratic equation.
Discriminant D = b^{2} - 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
x^{2} + (x + 2)^{2} = 52

Expand (x + 2)^{2}, group like terms and write the above equation with the right side equal to zero.
2x^{2} + 4x - 48 = 0

Multiply all terms in the above equation by 1/2 to obtain the following equivalent equation.
x^{2} + 2 x - 24 = 0

Find the discriminant of the above quadratic equation.
Discriminant D = b^{2} - 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.