|
-
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
x2 + 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
x2 + (x + 1)2 = 145
Expand and group like terms in the above equation and write it in standard form.
2 x2 + 2x - 144 = 0
Divide all terms in the above equation by 2.
x2 + 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.
x2 + 3x + 2 = 42
Factor and solve.
x2 + 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.
y2 + (y + 3)2 = (y + 6)2
Expand and group.
y2 + y2 + 6y + 9 = y2 + 12y + 36
y2 - 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 t2 + 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 t2 + 64 t + 32 = 80
Rewrite equation in standard form
-16 t2 + 64 t + 32 - 80 = 0
-16 t2 + 64 t - 48 = 0
Factor and solve
-16 (t2 - 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 - W2 = 96
20 W - W2 - 96 = 0
W2 - 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.
Solution
Let b be the base and b + 3 be the corresponding height. Area of a triangle is
54 = (1/2)× base × height = (1/2) b (b + 3)
Hence the equation
54 = (1/2) b (b + 3)
It is a quadratic equation that can be written as
b2 + 3 b - 108 = 0
Solve the above equation to find the solutions
b = 9 and b = - 12
The base must be positive and therefore it is equal to 9
The height is 3 feet longer than the base, hence
height = 9 + 3 = 12
-
The product of the first and the third of three consecutive positive integers is equal to 1 subtracted from the square of the second of these integers. Find the three integers.
Solution
Let x, x + 1 and x + 2 be the 3 consecutive numbers. The product of the first and the third is
x(x + 2)
1 subtracted from the square of the second is written as
(x + 1)2 - 1
Simplify both expressions
x(x + 2) = x 2 + 2 x and (x + 1)2 - 1 = x 2 + 2x + 1 - 1 = x 2 + 2x
The two expressions are equal for any real value x. Hence this problem has an infinite number of solutions and any set of 3 consecutive number is a solution to the given problem.
-
The product of two positive numbers is equal to 2 and their difference is equal to 7/2. Find the two numbers.
Solution
Let x be the smallest of the two numbers. The largest number may be written as
x + 7/2
The product of x and x + 7/2 is equal to 2
x(x + 7/2) = 2
The above equation may be written as
x2 + (7/2)x - 2 = 0
The above equation has two solutions
x = 1/2 and x = -4
We are looking for positive solutions, hence the two numbers are
x = 1/2
and
x + 7/2 = 1/2+7/2 = 4
-
The sum of the squares of three consecutive integers is equal to 77. What are the three integers?
Solution
Let x, x + 1 and x + 2 be the 3 consecutive integers. The sum of their squares is equal to 77, hence
x2 + (x + 1)2 + (x + 2)2 = 77
Expand the left side and simplify to rewrite the equation as
3 x2 + 6 x - 72 = 0
Solve to obtain two solutions
x = 4 and x = - 6
for x = 4, the integers are
4 , 5 and 6
for x = - 6, the integers are
-6 , - 5 and - 4
|
