This page presents a collection of Grade 8 quadratic equations word problems, complete with detailed step-by-step solutions and clear explanations. Expand the hidden solutions beneath each question to reveal the algebraic reasoning required to translate real-world scenarios into mathematical models.
The questions on this page help students master key applications of quadratic equations, including:
Two consecutive integers are of the form:
\[ x \quad \text{and} \quad x + 1 \]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 expanded and written in standard form as follows:
\[ x^{2} + x - 56 = 0 \]Factor and solve:
\[ (x - 7)(x + 8) = 0 \]Solutions: \( x = 7 \) or \( x = -8 \).
\( x = -8 \) is not valid since the problem states the numbers must be positive. Hence, \( x = 7 \).
The two consecutive positive integers are 7 and 8 (since \( 7 + 1 = 8 \)).
Two consecutive integers are of the form:
\[ x \quad \text{and} \quad x + 1 \]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:
\[ x^{2} + (x^{2} + 2x + 1) = 145 \] \[ 2x^{2} + 2x + 1 - 145 = 0 \] \[ 2x^{2} + 2x - 144 = 0 \]Divide all terms by 2 to simplify the equation:
\[ x^{2} + x - 72 = 0 \]Factor and solve:
\[ (x + 9)(x - 8) = 0 \]Solutions: \( x = -9 \) or \( x = 8 \).
Since the problem specifies positive integers, only \( x = 8 \) is valid.
The two consecutive positive numbers are 8 and 9.
The area of a rectangle equals length times width, so:
\[ (x + 2)(x + 1) = 42 \]Expand and group like terms:
\[ x^{2} + 3x + 2 = 42 \]Rewrite in standard form by subtracting 42 from both sides:
\[ x^{2} + 3x - 40 = 0 \]Factor and solve:
\[ (x + 8)(x - 5) = 0 \]Solutions: \( x = -8 \) and \( x = 5 \).
If \( x = -8 \), the width (\( x+1 \)) would be -7, which is physically impossible. Only \( x = 5 \) gives a valid positive length and width.
Calculate the dimensions:
The perimeter is \( 2 \times \text{length} + 2 \times \text{width} \):
\[ 2(7) + 2(6) = 14 + 12 = 26 \]The perimeter of the garden is 26.
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 (\( a^2 + b^2 = c^2 \)):
\[ y^{2} + (y + 3)^{2} = (y + 6)^{2} \]Expand the squares and simplify:
\[ y^{2} + (y^{2} + 6y + 9) = y^{2} + 12y + 36 \]Subtract \( y^2 + 12y + 36 \) from both sides to form a standard quadratic equation:
\[ y^{2} - 6y - 27 = 0 \]Factor and solve:
\[ (y - 9)(y + 3) = 0 \]Solutions: \( y = 9 \) or \( y = -3 \).
Only \( y = 9 \) is valid since length must be positive.
The length of the hypotenuse is \( y + 6 \):
\[ 9 + 6 = 15 \text{ cm} \]The object is 80 feet above ground when \( h = 80 \), so substitute 80 into the equation:
\[ -16t^{2} + 64t + 32 = 80 \]Rewrite in standard form by subtracting 80 from both sides:
\[ -16t^{2} + 64t + 32 - 80 = 0 \] \[ -16t^{2} + 64t - 48 = 0 \]Divide the entire equation by -16 to simplify:
\[ t^{2} - 4t + 3 = 0 \]Factor and solve:
\[ (t - 1)(t - 3) = 0 \]Solutions: \( t = 1 \) second and \( t = 3 \) seconds.
Explanation: The object reaches 80 feet on its way up at \( t = 1 \) second, continues to its peak, and then passes 80 feet again on its way back down at \( t = 3 \) seconds.
Let \( L \) be the length and \( W \) be the width. Based on the area:
\[ L \times W = 96 \]The perimeter is 40, which yields:
\[ 2L + 2W = 40 \]Divide by 2 and solve for \( L \):
\[ L + W = 20 \implies L = 20 - W \]Substitute this expression for \( L \) into the area equation:
\[ (20 - W) \times W = 96 \]Expand and rearrange into standard quadratic form:
\[ 20W - W^{2} = 96 \implies W^{2} - 20W + 96 = 0 \]Factor and solve:
\[ (W - 8)(W - 12) = 0 \]Solutions: \( W = 8 \) or \( W = 12 \).
Now, find the corresponding lengths:
By convention, length is typically the longer side. Therefore, the dimensions are width = 8 m and length = 12 m.
Let \( b \) be the base. Then the height is \( b + 3 \). Using the area formula for a triangle (\( A = \frac{1}{2} \times \text{base} \times \text{height} \)):
\[ 54 = \frac{1}{2} \times b \times (b + 3) \]Multiply both sides by 2 to clear the fraction:
\[ 108 = b(b + 3) \]Expand and rewrite as a standard quadratic equation:
\[ b^{2} + 3b - 108 = 0 \]Factor and solve:
\[ (b + 12)(b - 9) = 0 \]Solutions: \( b = -12 \) or \( b = 9 \).
Since a base length must be positive, \( b = 9 \).
The height is \( b + 3 = 9 + 3 = 12 \).
The base is 9 feet and the height is 12 feet.
Let the three consecutive positive integers be \( x \), \( x + 1 \), and \( x + 2 \).
The product of the first and third is:
\[ x(x + 2) = x^{2} + 2x \]One less than the square of the second integer is:
\[ (x + 1)^{2} - 1 = (x^{2} + 2x + 1) - 1 = x^{2} + 2x \]Notice that both algebraic expressions simplify to \( x^2 + 2x \). This creates the equation:
\[ x^{2} + 2x = x^{2} + 2x \]Since this results in \( 0 = 0 \), it is an algebraic identity. This means the equation holds true for all real numbers \( x \).
Therefore, any set of three consecutive positive integers satisfies this condition (e.g., 1, 2, 3 or 10, 11, 12).
Let \( x \) be the smaller number. Since their difference is \( \frac{7}{2} \), the larger number is \( x + \frac{7}{2} \).
Their product is 2:
\[ x \left( x + \frac{7}{2} \right) = 2 \]Expand the expression:
\[ x^{2} + \frac{7}{2} x = 2 \]Rewrite as a standard quadratic equation by subtracting 2:
\[ x^{2} + \frac{7}{2} x - 2 = 0 \]Multiply the entire equation by 2 to clear the fraction:
\[ 2x^{2} + 7x - 4 = 0 \]Factor the quadratic equation:
\[ (2x - 1)(x + 4) = 0 \]Solutions: \( x = \frac{1}{2} \) or \( x = -4 \).
The problem asks for positive numbers, so the smaller number is \( x = \frac{1}{2} \).
The larger number is \( \frac{1}{2} + \frac{7}{2} = \frac{8}{2} = 4 \).
The two numbers are \( \frac{1}{2} \) and 4.
Let the three consecutive integers be \( x \), \( x + 1 \), and \( x + 2 \).
The sum of their squares is 77:
\[ x^{2} + (x + 1)^{2} + (x + 2)^{2} = 77 \]Expand the squares:
\[ x^{2} + (x^{2} + 2x + 1) + (x^{2} + 4x + 4) = 77 \]Combine like terms:
\[ 3x^{2} + 6x + 5 = 77 \]Subtract 77 from both sides to form a standard quadratic equation:
\[ 3x^{2} + 6x - 72 = 0 \]Divide the entire equation by 3 to simplify:
\[ x^{2} + 2x - 24 = 0 \]Factor and solve:
\[ (x + 6)(x - 4) = 0 \]Solutions: \( x = -6 \) or \( x = 4 \).
Because the problem asks for integers (and does not restrict them to just positive ones), there are two valid sets of answers.