Explore Grade 11 algebra questions covering functions, polynomials, circle and parabola equations, and simplifying expressions. Each topic includes step-by-step solutions designed to help students understand and master key algebraic concepts. Perfect for revision, practice, and building confidence in problem-solving.
Complete the square in the quadratic function \( f \) given by \[ f(x) = 2 x^2 - 6x + 4 \]
Factor out 2 from the first two terms: \[ f(x) = 2(x^2 - 3x) + 4 \] Add and subtract \(\left(\dfrac{-3}{2}\right)^2\) inside the parentheses: \[ = 2\left(x^2 - 3x + \left(\dfrac{-3}{2}\right)^2 - \left(\dfrac{-3}{2}\right)^2\right) + 4 \] Write \( x^2 - 3x + \left(\dfrac{3}{2}\right)^2 \) as a square \( \left(x - \dfrac{3}{2}\right)^2 \) : \[ = 2 \left(\left(x - \dfrac{3}{2}\right)^2 - \left(\dfrac{-3}{2}\right)^2 \right) + 4 \] \[ = 2 \left(x - \dfrac{3}{2}\right)^2 - 2 \left(\dfrac{-3}{2}\right)^2 + 4 \] \[ = 2 \left(x - \dfrac{3}{2}\right)^2 - \dfrac{9}{2} + 4 \] Group \( - \dfrac{9}{2} + 4 \) \[ = 2\left(x - \dfrac{3}{2}\right)^2 - \dfrac{1}{2} \]
Find the point(s) of intersection of the parabola with equation \( y = x^2 - 5 x + 4 \) and the line with equation \( y = 2 x - 2 \)
The coordinates of the points of intersection of the parabola and the line are found by solving the system of the equations of the parabola and the line: \[ \begin{cases} y = x^2 - 5x + 4 \\\\ y = 2x - 2 \end{cases} \] Substitute \( y \) by \( 2x - 2 \) in the first equation: \[ 2x - 2 = x^2 - 5x + 4 \] Write the quadratic equations in standard form (one side equal to zero) \[ x^2 - 7 x + 6 = 0 \] Solution of the quadratic equation: \[ x = 1 \quad \text{and} \quad x = 6 \] Find the \( y \) coordinates:
For \( x = 1 \) , \[ y = 2x - 2 = 2(1) - 2 = -1 \]
For \( x = 6 \) , \[ y = 2 x - 2 = 2(6) - 2 = 10 \] Points of intersection: \[ (1, 0) \quad \text{and} \quad (6, 10) \]
Find the constant \( k \) so that : \[ -x^2 - (k + 7)x - 8 = -(x - 2)(x - 4) \] for all real values of \( x \)
Given equation: \[ - x^2 - (k + 7)x - 8 = -(x - 2)(x - 4) \] Expanding the right-hand side: \[ - x^2 - (k + 7)x - 8 = -x^2 + 6x - 8 \] Two polynomials are equal if their corresponding coefficients are equal, hence: \[ -(k + 7) = 6 \] Solving for \( k \): \[ k = -13 \]
Find the center and radius of the circle with equation \[ x^2 + y^2 -2x + 4y - 11 = 0 \]
Put terms in \(x\) together and terms in \(y\) together: \[ (x^2 - 2x) + (y^2 + 4y) = 11 \] Complete the square in \( x^2 - 2x \) as follows \[ (x^2 - 2x) = (x - 1)^2 - 1 \] Complete the square in \( y^2 + 4y \) as follows \[ y^2 + 4y = (y + 2)^2 - 4 \] Rewrite the given equation as \[ (x - 1)^2 - 1 + (y + 2)^2 - 4 = 11 \] Write the equation of the circle in standard form \( (x - h)^2 + (y - k)^2 = r^2 \): \[ (x - 1)^2 + (y + 2)^2 = 4^2 \] Identify the center \( (h,k) \) and radius \(r\): \[ x - h = x - 1 \quad \text{gives} \quad h = 1 \] \[ y - k = y + 2 \quad \text{gives} \quad k = -2 \] \[ r^2 = 4^2 \quad \text{gives} \quad r = 4 \] Hence \[ \text{Center is at: } (1, -2), \quad \text{Radius = } 4 \]
Find the constant \( k \) so that the quadratic equation \[ 2 x^2 + 5 x - k = 0 \] has two real solutions.
Given quadratic equation: \[ 2x^2 + 5x - k = 0 \] Calculate the discriminant: \[ \Delta = 5^2 - 4(2)(-k) = 25 + 8k \] A quadratic equation has two real solutions when the discriminant is positive: \[ 25 + 8 k > 0 \] Solve for \( k \): \[ k > -\frac{25}{8} \]
Find the constant \( k \) so that the system of the two equations: \[ 2x + ky = 2 \quad \text{and} \quad 5x - 3y = 7 \] has no solutions.
According to Cramer's rule, if the determinant \( D \) of the coefficient matrix to equal to zero and one of the determinants \( D_x \) or \( D_y \) is not equal to zero, the system has no solutions.
Compute the Determinant of the Coefficient Matrix of the the system: \[ \begin{aligned} 2x + ky &= 2 \\ 5x - 3y &= 7 \end{aligned} \] The coefficient matrix is: \[ A = \begin{bmatrix} 2 & k \\ 5 & -3 \end{bmatrix} \] Its determinant: \[ D = \begin{vmatrix} 2 & k \\ 5 & -3 \end{vmatrix} = (2)(-3) - (5)(k) = -6 - 5k \] Setting \( D = 0 \): \[ -6 - 5k = 0 \] \[ k = -\frac{6}{5} \] We define the determinants \( D_x \) and \( D_y \) by replacing the corresponding column in \( A \) with the constants matrix.
Determinant \( D_x \) is defined by replacing the first column of \( A \) with the constants \( \begin{bmatrix} 2 \\ 7 \end{bmatrix} \): \[ D_x = \begin{vmatrix} 2 & k \\ 7 & -3 \end{vmatrix} \] \[ D_x = (2)(-3) - (7)(k) = -6 - 7k \] Substituting \( k = -\frac{6}{5} \): \[ D_x = -6 - 7\left(-\frac{6}{5}\right) \] \[ D_x = -6 + \frac{42}{5} \] \[ D_x = \frac{-30 + 42}{5} = \frac{12}{5} \neq 0 \] Since the coefficient determinant \( D = 0 \) but \( D_x \neq 0 \), the system is inconsistent and has no solutions when : \[ k = -\frac{6}{5} \]
Factor the expression \[ 6 x^2 - 13 x + 5 \]
We will factor the quadratic expression step by step: \[ 6x^2 - 13x + 5 \] In the quadratic expression \( ax^2 + bx + c \), we identify: \[ a = 6 \] \[ b = -13 \] \[ c = 5 \] Multiply \( a \) and \( c \): \[ 6 \times 5 = 30 \] We need two numbers that:
1) Multiply to 30
2) Add to -13
The numbers \( -10 \) and \( -3 \) satisfy these conditions: \[ -10 \times -3 = 30 \] \[ -10 + (-3) = -13 \] We use the numbers numbers \( -10 \) and \( -3 \) to split \( -13x \) as \( -10x - 3x \): \[ 6x^2 - 10x - 3x + 5 \] Group the terms: \[ (6x^2 - 10x) + (-3x + 5) \] Factor out the common terms in each group: \[ 2x(3x - 5) - 1(3x - 5) \] Since \( (3x - 5) \) is common, the given expression is factored as: \[ 6x^2 - 13x + 5 = (2x - 1)(3x - 5) \]
Simplify \( i^{231} \) where \( i \) is the imaginary unit and is defined as: \[ i = \sqrt{-1} \]
Note that: \[ i^4 = 1 \] Also, observe that: \[ 231 = 4 \times 57 + 3 \] Hence: \[ i^{231} = i^{4 \times 57 + 3} = i^{4 \times 57} \; i^3 = (i^4)^{57} \times i^3 \] Since \( i^4 = 1 \), we get: \[ = 1^{57} \times (-i) = -i \] Hence \[ i^{231} = - 1 \]
What is the remainder when \( f(x) = (x - 2)^{54} \) is divided by \( x - 1\) ?
According to the remainder theorem, the remainder \( r \) of the division of \( f(x) = (x - 2)^{54} \) by \( x - 1\) is given by: \[ r = f(1) = (1 - 2)^{54} = (-1)^{54} = 1 \]
Find \( b \) and \( c \) so that the parabola with equation \( y = 4 x^2 - b x - c \) has a vertex with coordinates \( (2 , 4) \) ?
The formula for the x-coordinate of the vertex: \[ h = \frac{- (-b)}{2\cdot4} = \frac{b}{8} = 2 \] Solve for \( b \): \[ b = 16 \] The vertex point is a solution to the equation of the parabola: \[ 4(2)^2 - 16(2) - c = 4 \] Solve for \( c \): \[ c = -20 \] Hence \[ b = 16 \; , \; c = -20 \]
Find all zeros of the polynomial \[ P(x) = x^3 - 3 x^2 - 10 x + 24 \] knowing that \( x = 2 \) is a zero of \(P(x) \).
Since \( x = 2 \) is a root, \( (x - 2) \) is a factor of \( P(x) \).
Use polynomial division to divide \( P(x) \) by \( (x - 2) \) \[ \begin{align*} P(x) &= x^3 - 3x^2 - 10x + 24 \\ &= (x - 2) Q(x) \end{align*} \] Divide \( P(x) \) by \( (x - 2) \) using long division or synthetic division to obtain: \[ Q(x) = \dfrac{x^3 - 3x^2 - 10x + 24 }{x - 2} = x^2 - x - 12 \] So we now have: \[ P(x) = (x - 2)(x^2 - x - 12) \] Factor the quadratic \[ x^2 - x - 12 = (x - 4)(x + 3) \] Final Factorization: \[ P(x) = (x - 2)(x - 4)(x + 3) \] Use the three factors to list all the zeros \[ x = 2, \quad x = 4, \quad x = -3 \]
If \( x \) is an integer, what is the greatest value of \( x \) which satisfies the inequality \[ 5 \lt 2x + 2 \lt 9? \]
Given: \[ 5 \lt 2x + 2 \lt 9 \] Solve the compound inequality: \[ \frac{3}{2} \lt x \lt \frac{7}{2} \] Since \( \frac{7}{2} = 3.5 \), the greatest integer value of \( x \) which satisfies the given inequality is 3 (the largest integer less than \( \frac{7}{2} \))
Sets \( A \) and \( B \) are given by: \[ A = \{2 , 3 , 6 , 8, 10\} \; , \; B = \{3 , 5 , 7 , 9\} \].
a) Find the intersection of the sets \( A \) and \( B \).
b) Find the union of the sets \( A \) and \( B \).
a) The common element to both \( A \) and \( B \) is 3, hence: \[ A \cap B = \{3\} \] b) All elements of \( A \) and \( B \) are in the union. Elements that are common to both \( A \) and \( B \) are listed only once since it is a set, hence: \[ A \cup B = \{2, 3, 6, 8, 10, 5, 7, 9\} \]
Simplify \[ \left| - x^2 + 4x - 4 \right| \]
Given: \[ | -x^2 + 4x - 4 | \] Rewrite the expression by factoring out the negative sign: \[ = | - (x^2 - 4x + 4) | \] Complete the square: \[ = | -(x - 2)^2 | \] Use the property that the absolute value of a negative square is just the square: \[ = (x - 2)^2 \]
Find the constant \( k \) so that the line with equation \( y = k x \) is tangent to the circle with equation \[ (x - 3)^2 + (y - 5)^2 = 4 \].
A line and a circle are tangent if they have one point of intersction only which is the point of tangency.
Given \[ (x - 3)^2 + (y - 5)^2 = 4 \] Substitute \( y \) by \( k x \): \[ (x - 3)^2 + (k x - 5)^2 = 4 \] Expand \[ x^2 - 6 x + 9 + ( k x )^2 -10 kx + 25 = 4 \] Write the quadratic equation in standard form: \[ x^2(1 + k^2) - x (6 + 10k) + 30 = 0 \] For the line \( y = k x \) to be tangent to the circle, the discriminant \( \Delta \)of the above quadratic equation must be equal to zero: \[ \Delta = b^2 - 4 a c = (-(6 + 10k))^2 - 4(1 + k^2)(30) = 0 \] Expand the above equation: \[ -20 k^2+120 k-84 = 0 \] Solve the above quadratic equation to obatin the values of \( k \) for the line is tangent to the circle: \[ k = \frac{15+2\sqrt{30}}{5} \quad \text{or} \quad k = \frac{15-2\sqrt{30}}{5} \]