Quadratic Equations Problems
with Solutions and Explanations for Grade 8
Solutions with full explanations to grade 8 questions on quadratic
equations. Some of these problems may be challenging and are therefore worth solving even if they take time.
We learn by solving problems that we do not know how to solve at first.

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.
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
b^{2} + 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
x^{2} + (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
x^{2} + (x + 1)^{2} + (x + 2)^{2} = 77
Expand the left side and simplify to rewrite the equation as
3 x^{2} + 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

More References and Links
More Middle School Math (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
More High School Math (Grades 10, 11 and 12)  Free Questions and Problems With Answers
More High School Math (Grades 10, 11 and 12)  Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Home Page