This page presents a collection of Grade 8 quadratic equations problems, complete with detailed solutions and clear explanations. These word problems help students practice and master key concepts in quadratic equations. For more on the topic, visit our quadratic equations tutorial.
The product of two positive consecutive integers is equal to 56. Find the two integers.
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 numbers must be positive. Hence
\( x = 7 \) and \( x + 1 = 8 \) are the two consecutive numbers.
The sum of the squares of two consecutive numbers is equal to 145. Find the two numbers.
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, then write in standard form
\[ 2x^{2} + 2x - 144 = 0 \]Divide all terms by 2
\[ x^{2} + x - 72 = 0 \]Factor and solve
\[ (x + 9)(x - 8) = 0 \]Solutions: \( x = 8 \) (only positive solution)
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.
Area equals length times width, so
\[ (x + 2)(x + 1) = 42 \]Expand and group like terms
\[ x^{2} + 3x + 2 = 42 \]Rewrite in standard form
\[ x^{2} + 3x - 40 = 0 \]Factor and solve
\[ (x + 8)(x - 5) = 0 \]Solutions: \( x = -8 \) and \( x = 5 \)
Only \( x = 5 \) gives positive length and width
Length: \( x + 2 = 7 \)
Width: \( x + 1 = 6 \)
Perimeter is
\[ 2 \times \text{length} + 2 \times \text{width} = 14 + 12 = 26 \]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?
Let \( y \) be the length of the shorter leg. Then the longer leg is
\( y + 3 \)
The hypotenuse is 3 cm longer than the longer leg, so
\( (y + 3) + 3 = y + 6 \)
Use the Pythagorean theorem
\[ y^{2} + (y + 3)^{2} = (y + 6)^{2} \]Expand and simplify
\[ y^{2} + y^{2} + 6y + 9 = y^{2} + 12y + 36 \] \[ y^{2} - 6y - 27 = 0 \]Factor and solve
\[ (y - 9)(y + 3) = 0 \]Only \( y = 9 \) is valid since length must be positive.
Length of hypotenuse:
\( y + 6 = 15 \text{ cm} \)
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?
The object is 80 feet above ground when \( h = 80 \), so
\[ -16t^{2} + 64t + 32 = 80 \]Rewrite in standard form
\[ -16t^{2} + 64t + 32 - 80 = 0 \implies -16t^{2} + 64t - 48 = 0 \]Factor and solve
\[ -16(t^{2} - 4t + 3) = 0 \] \[ -16 (t - 1)(t - 3) = 0 \]Solutions: \( t = 1 \) second and \( t = 3 \) seconds.
The object reaches 80 feet at \( t = 1 \), goes up, then comes down and again passes 80 feet at \( t = 3 \) before falling further.
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.
Let \( L \) be the length and \( W \) be the width. Given
\[ L \times W = 96 \]Perimeter is 40, so
\[ 2L + 2W = 40 \implies L + W = 20 \implies L = 20 - W \]Substitute into area equation
\[ (20 - W) \times W = 96 \]Expand and rearrange
\[ 20W - W^{2} = 96 \implies W^{2} - 20W + 96 = 0 \]Factor and solve
\[ (W - 8)(W - 12) = 0 \]Solutions: \( W = 8 \), \( W = 12 \)
Find corresponding \( L \)
\[ \begin{cases} W = 8 \implies L = 12 \\ W = 12 \implies L = 8 \end{cases} \]Assuming length is longer, the 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.
Let \( b \) be the base, then the height is \( b + 3 \). Area formula:
\[ 54 = \frac{1}{2} \times b \times (b + 3) \]Multiply both sides by 2:
\[ 108 = b(b + 3) \]Rewrite as quadratic:
\[ b^{2} + 3b - 108 = 0 \]Solve the quadratic:
\[ b = 9 \quad \text{or} \quad b = -12 \]Base must be positive, so \( b = 9 \). Height is
\[ 9 + 3 = 12 \]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 numbers. Find the three integers.
Let the integers be \( x \), \( x + 1 \), and \( x + 2 \).
Product of first and third is:
\[ A = x(x + 2) = x^{2} + 2x \]One less than the square of the second:
\[ B = (x + 1)^{2} - 1 = x^{2} + 2x + 1 - 1 = x^{2} + 2x \]Both expressions \( A \) and \( B \) are equal for all real \( x \). Therefore, any set of three consecutive integers is such that: product of the first and the third of three consecutive positive integers is equal to 1 subtracted from the square of the second
The product of two positive numbers is equal to 2 and their difference is equal to \( \dfrac{7}{2} \). Find the two numbers.
Let \( x \) be the smaller number. Then the larger number is \( x + \frac{7}{2} \).
The product is:
\[ x \left( x + \frac{7}{2} \right) = 2 \]Expand and write as a standard quadratic equation:
\[ x^{2} + \frac{7}{2} x - 2 = 0 \]Solve the equation:
\[ x = \frac{1}{2} \quad \text{or} \quad x = -4 \]Positive solution is \( x = \frac{1}{2} \), so the numbers are
\[ \frac{1}{2} \quad \text{and} \quad \frac{1}{2} + \frac{7}{2} = 4 \]The sum of the squares of three consecutive integers is equal to 77. What are the three integers?
Let the integers be \( x \), \( x + 1 \), and \( x + 2 \).
The sum of their squares is
\[ x^{2} + (x + 1)^{2} + (x + 2)^{2} = 77 \]Expand and simplify:
\[ 3x^{2} + 6x - 72 = 0 \]Solve the quadratic:
\[ x = 4 \quad \text{or} \quad x = -6 \]For \( x = 4 \), the integers are
\( 4, 5, 6 \)
For \( x = -6 \), the integers are
\( -6, -5, -4 \)