

The product of two positive consecutive integers is equal to 56. Find the two integers.
Solution
Two consecutive integers are of the form
x and x + 1
Their product is equal to 56
x(x + 1) = 56
Solve and find the two numbers x and x + 1. The above equation may be written as follows
x^{2} + x  56 = 0
Factor and solve
(x  7)(x + 8) = 0
solutions: x = 7 , x = 8
x=  8 is not valid since the number sought must be positive. Hence
x = 7 and x + 1 = 8 are the two consecutive numbers.

The sum of the squares of two consecutive positive integers is equal to 145. Find the two numbers.
Solution
Two consecutive integers are of the form
x and x + 1
The sum of their squares is equal to 145
x^{2} + (x + 1)^{2} = 145
Expand and group like terms in the above equation and write it in standard form.
2 x^{2} + 2x  144 = 0
Divide all terms in the above equation by 2.
x^{2} + x  72 = 0
Factor and solve.
(x + 9)(x  8) = 0
Solutions: x = 8 is the only solution to our problem since we are looking for positive integers.
The two consecutive numbers are.
x = 8 and x + 1 = 9

A rectangular garden has length of x + 2 and width of x + 1 and an area of 42. Find the perimeter of this garden
Solution
Area is equal to length times width. Hence
(x + 2)(x + 1) = 42
Expand and group like terms.
x^{2} + 3x + 2 = 42
Factor and solve.
x^{2} + 3x  40 = 0
(x + 8)(x  5) = 0
Solutions: x = 8 and x = 5
The only solution that gives positive length and width is x = 5. Hence
length = x + 2 = 5 + 2 = 7
width = x + 1 = 6
Perimeter is equal to
2 × length + 2 × width = 14 + 12 = 26

A right triangle has one leg 3 cm longer that the other leg. Its hypotenuse is 3 cm longer than its longer leg. What is the length of the hypotenuse?
Solution
Let y be the length of the shorter leg. Hence
y + 3 is the length of the longer leg.
The hypotenuse is 3 cm longer than the longer leg. Hence the length of the hypotenuse is equal to
(y + 3) + 3 = y + 6
We now use Pythagora's theorem.
y^{2} + (y + 3)^{2} = (y + 6)^{2}
Expand and group.
y^{2} + y^{2} + 6y + 9 = y^{2} + 12y + 36
y^{2}  6y  27 = 0
Factor and solve for y.
(y  9)(y + 3) = 0
Solution
y = 9 is the only acceptable solution since the dimensions of the trangle musty be positive.
Length of hypotenuse
y + 6 = 9 + 6 = 15 cm

The height h above the ground of an object propelled vertically is given by h = 16 t^{2} + 64 t + 32 , where h is in feet and t is in seconds. At what time t will the object be 80 feet above ground?
Solution
The object is 80 feet above ground when h = 80. Hence
16 t^{2} + 64 t + 32 = 80
Rewrite equation in standard form
16 t^{2} + 64 t + 32  80 = 0
16 t^{2} + 64 t  48 = 0
Factor and solve
16 (t^{2}  4 t + 3) = 0
16(t  1)(t  3) = 0
Two solutions
t = 1 second and t = 3 seconds
The object is propelled vertically will be 80 feet at t= 1 second, goes up then down and again is at 80 feet above ground than fall.

The area of a rectangle is equal to 96 square meters. Find the length and width of the rectangle if its perimeter is equal to 40 meters.
Solution
Let L be the length and W be the width of the rectangle. Area is 96; hence
L × W = 96
Perimeter is 40; hence
2(L + W) = 40 , or , L + W = 20 , or , L = 20  W
Substitute L by 20  W in equation L × W = 96
(20  W) × W = 96
Expand and group like terms
20 W  W^{2} = 96
20 W  W^{2}  96 = 0
W^{2}  20 W + 96 = 0
Factor and solve
(W  8)(W  12) = 0
Solutions: W = 8, W = 12
Find L for each W
W = 8 , L = 20  W = 20  8 = 12
W = 12 , L = 20  W = 20  12 = 8
Assuming length longer than width, the two dimensions are
W = 8 and L = 12

The height of a triangle is 3 feet longer than its corresponding base. The area of the triangle is equal to 54 square feet. Find the base and the height of the triangle.

The product of the first and the third of three consecutive positive numbers is equal to 1 subtracted from the square of the second of these numbers. Find the three numbers.

The product of two positive numbers is equal to 2 and their difference is equal to 7/2. Find the two numbers.

The sum of the squares of three consecutive integers is equal to 77. What are the three integers?

