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.
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.
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 \( x + 1 = 9 \).
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
\[ 2 \times \text{length} + 2 \times \text{width} = 14 + 12 = 26 \]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 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.
Let \( L \) be the length and \( W \) be the width. Given
\[ L \times W = 96 \]Perimeter is 40, so
\[ 2(L + W) = 40 \implies L + W = 20 \implies L = 20 - W \]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 \( L = 12 \)
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
\[ 9 + 3 = 12 \]Let the integers be \( x \), \( x + 1 \), and \( x + 2 \).
Product of first and third:
\[ x(x + 2) = x^{2} + 2x \]One less than the square of the second:
\[ (x + 1)^{2} - 1 = x^{2} + 2x + 1 - 1 = x^{2} + 2x \]Both expressions are equal for all real \( x \). Therefore, any set of three consecutive integers satisfies the condition.
Let \( x \) be the smaller number. Then the larger number is \( x + \frac{7}{2} \).
The product is:
\[ x \left( x + \frac{7}{2} \right) = 2 \]Rewrite as a quadratic:
\[ x^{2} + \frac{7}{2} x - 2 = 0 \]Solve the quadratic:
\[ x = \frac{1}{2} \quad \text{or} \quad x = -4 \]Positive solution is \( x = \frac{1}{2} \), so the numbers are
\[ \frac{1}{2} \quad \text{and} \quad \frac{1}{2} + \frac{7}{2} = 4 \]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
\( -6, -5, -4 \)