Grade 12 Algebra questions and problems, complete with step-by-step solutions, are provided here. Some of these problems may be challenging, requiring time and effort to solve. However, tackling difficult questions helps you develop critical thinking and problem-solving skills.
Working in groups on these problems is a great way to collaborate, share ideas, and learn from each other. If you discover alternative solutions to any of the questions, feel free to share them!
Find all rational zeros of \( P(x) = x^3 - 7 x + 6 \) and factor \(P(x)\)
The leading coefficient is \( 1 \), and its factors are:
\[ \pm 1 \]The constant term is \( 6 \), and its factors are:
\[ \pm 1, \pm 2, \pm 3, \pm 6 \] - The possible rational zeros are: \[ \pm 1, \pm 2, \pm 3, \pm 6 \]Testing for Rational Zeros, using direct substitution and calculations:
\[ P(1) = 0, \quad P(2) = 0, \quad P(-3) = 0 \]Since these values satisfy \( P(x) = 0 \), they are the zeros of \( P(x) \). The polynomial \( P(x) \) has the following rational zeros: \[ x = 1, \quad x = 2, \quad x = -3 \]
Thus, the polynomial can be factored further using these roots.
\[ P(x) = (x-1)(x-2)(x+3) \)Round all real zeros in the graph to the nearest integer and find a polynomial function \(P\) of lowest degree, with the absolute value of the leading coefficient equal to 1, that has the indicated graph.
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From the graph, \(x = -3 \) is a zero of multiplicity \(2\), \(x = 0\) is a zero of multiplicity 1 and \(x = 2\) is a zero of multiplicity 2.
The polynomial has degree 5 and its graph is falling on the right and rising on the left therefore the leading coefficient is negative and equal to \(-1\), hence \( P(x) \) is given by
\[ P(x) = - x (x + 3)^2 (x - 2)^2 \]Find \( a, b, \) and \( c \) so that the graph of the quadratic function \[ f(x) = ax^2 + bx + c \] has a vertex at \( (-2,1) \) and passes through the point \( (0,-3) \).
Use the Given Point \( (0, -3) \): Substituting \( x = 0 \) and \( f(0) = -3 \): \[ -3 = a(0 + 2)^2 + 1 \] \[ -3 = 4a + 1 \] Solving for \( a \): \[ 4a = -4 \] \[ a = -1. \] Substituting \( a = -1 \) into the vertex form: \[ f(x) = -(x + 2)^2 + 1 \] Expanding: \[ f(x) = -x^2 - 4x - 3. \] Comparing with the standard quadratic form \( f(x) = ax^2 + bx + c \), we find: \[ a = -1, \quad b = -4, \quad c = -3. \]
\( f(x) \) is a quadratic function such that \( f(1) = 3 \) and \( f(5) = 3 \). Find the \( x \) coordinate of the vertex of the graph of \( f \).
Since \( f(x) \) is a quadratic function such that \( f(1) = 3 \) and \( f(5) = 3 \), the two points \( (1, 3) \) and \( (5, 3) \) lie on the parabola, and since the function is quadratic, the x coordinate of the vertex is equal to x coordinte of the midpoint of these two points since they have the same y coordinte \( 3 \).
The \( x \)-coordinate of the midpoint is: \[ x_{\text{vertex}} = \dfrac{1 + 5}{2} = \dfrac{6}{2} = 3. \] The \( x \)-coordinate of the vertex of the graph of \( f(x) \) is \( \boxed{3} \).
Find \(a \) and \( b \) so that the rational function given by: \[ f(x) = \dfrac{a x^4 + b x^3 + 3}{x^3 - 2} \] has an oblique asymptote given by: \( \; y = 2x - 3 \)
The oblique asymptote is the quotient resulting from the long division of \( a x^4 + b x^3 + 3 \) by \( x^3 - 2 \).
Using long division, we obtain
\[ \dfrac{ax^4 + bx^3 + 3}{x^3 - 2} = ax + b + \dfrac{2ax + (3 + 2b)}{x^3 - 2} \] The quotient \( ax + b \) is the oblique asymptote and it must be equal to \( 2x - 3 \). \[ ax + b = 2x - 3 \] Since two polynomials are equal if and only if their corresponding coefficients are equal, we compare and obtain: \[ a = 2, \quad b = -3 \]Given \[ \log_9 (x^3) = \log_2(8) \] Simplify right-hand side of the given equation \[ \log_2(8) = \log_2 (2^3) = 3 \] Rewrite the above equation \[ \log_9 (x^3) = 3 \] Rewrite 3 using a logarithm with base 9 \[ \log_9 (x^3) = \log_9(9^3) \] Write the algebraic equation from the previous equation \[ x^3 = 9^3 \] Solve for \(x\) \[ x = 9 \quad \]
Find the value of \( \log_y (x^4) \) if \( \log_x (y^3) = 2 \).
Solve for \(x\) the equation \[ \log_x (8e^3) = 3 \]
Given \[ a^2 - b^2 = 8 \] Square both sides and expand \[ a^4 + b^4 - 2a^2b^2 = 8^2 \quad (I) \] Given \[ a b = 2 \] Square both sides of the above \[ (a b)^2 = a^2 b^2 = 2^2 =4 \] Substitute in (I) \[ a^4 + b^4 - 2(4) = 8^2 \] Hence \[ a^4 + b^4 = 72 \]
What are the maximum value and minimum values of the function \[ f(x) = \left|2 \sin \left(2 x - \dfrac{\pi}{3}\right) - 5\right| + 3 \]
The range of a sine function is
\[ -1 \leq \sin\left( 2x - \dfrac{\pi}{3} \right) \leq 1 \]multiply all terms of the double inequality by 2
\[ -2 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) \leq 2 \]Add -5 to all terms of the inequality \[ -2 - 5 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \leq 2 - 5 \]
Simplify \[ -7 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \leq -3 \]
Change the above inequality using absolute value \[ 3 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| \leq 7 \]
Add 3 to all terms of the double inequality \[ 3 + 3 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| + 3 \leq 7 + 3 \]
Simplify \[ 6 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| + 3 \leq 10 \] The maximum value of \( f(x) \) is equal to \( 10 \) and the minimum value of \( f(x) \) is equal to \( 6 \).
If \(x \lt -7\), simplify \[ \left| 4 - |3 + x| \right| \].
If \( x \lt -7 \) then \[ x \lt - 3 \]
Add \( 3 \) to the above inequatlity and simplify \[ x + 3 \lt 0 \]
According to the definition of the absolute value \[ |3 + x| = -(3 + x) \]
The given expression simplifies to \[ |4 - |3 + x|| = |4 - (- (3 + x))| = |x + 7| \]
Since \( x + 7 \lt 0 \) \[ |x + 7| = - (x + 7) = - x - 7 \]
and hence \[ |4 - |3 + x|| = -x - 7 \]
A car travels from \( A \) to \( B \) at an average speed of 50 km/hour. At what average speed would it have to travel from \( B \) to \( A \) to average 60 km/hour for the whole trip?
Let \( d \) be the distance between \( A \) and \( B \) is \( d \). The time taken to travel from A to B is given by: \[ t_1 = \dfrac{d}{50} \] where \( 50 \) km/hour is the average speed from \( A \) to \( B \).
Let the average speed for the return trip (from \( B \) to \( A \)) be \( v \). The time taken for the return trip is: \[ t_2 = \dfrac{d}{v} \]
The total time for the entire trip is: \[ t_{\text{total}} = t_1 + t_2 = \dfrac{d}{50} + \dfrac{d}{v} \]
The total distance (away and return) is: \[ d + d = 2d \]
The average speed for the whole trip is given by: \[ \text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{2d}{t_{\text{total}}} \] We are told the average speed for the whole trip should be 60 km/hour, so: \[ \dfrac{2d}{\dfrac{d}{50} + \dfrac{d}{v}} = 60 \]
Cancel \( d \) from both the numerator and denominator: \[ \dfrac{2}{\dfrac{1}{50} + \dfrac{1}{v}} = 60 \]
Multiply both sides by \( \left( \dfrac{1}{50} + \dfrac{1}{v} \right) \): \[ 2 = 60 \left( \dfrac{1}{50} + \dfrac{1}{v} \right) \]
Simplify: \[ 2 = \dfrac{60}{50} + \dfrac{60}{v} \] \[ 2 = 1.2 + \dfrac{60}{v} \]
Subtract 1.2 from both sides: \[ 0.8 = \dfrac{60}{v} \]
Solve for \( v \): \[ v = \dfrac{60}{0.8} = 75 \]
Thus, the average speed for the return trip from \( B \) to \( A \) needs to be \( 75 \) km/hour to average \( 60 \) km/hour for the whole trip.
Find \(x\) and \( y \) if \( x^2 - y^2 = -12 \) and \( x + y = 6 \).
Substitute \( x + y \) by 6 in the above \[ 6(x - y) = -12 \]
Solve for \( x - y \) \[ (x - y) = -2 \] We now have a system of equations to solve \[ \begin{cases} x - y = -2 \\ x + y = 6 \end{cases} \]
Solve the system above \[ x = 2, \quad y = 4 \]
\( f(x) \) is a function such that \( f(x) + 3 f(8 - x) = x \) for all real numbers \( x \). Find the value of \( f(2) \).
Given \[ f(x) + 3f(8 - x) = x \]
Substitute \( x = 2\) in the given equation \[ f(2) + 3 f(6) = 2 \tag{A} \]
Substitute \( x = 6\) in the given equation \[ f(6) + 3 f(2) = 6 \tag{B}\]
Solve equation (B) for \( f(6) \): \[ f(6) = 6 - 3f(2) \tag{C} \]
Substitute into equation (A): \[ f(2) + 3(6 - 3f(2)) = 2 \]
Expand and group \[ - 8 f(2) + 18 = 2 \]
Solve for } f(2): \[ f(2) = 2 \]
Let \( f(x) \) be a function such that \[ f(2x + 1) = 2f(x) + 1 \quad \text{for all real numbers } x \] Find the value of \( f(3) \) given that \( f(0) = 2 \).
Find \( b \) so that the line with equation \( y = 2 x + b \) is tangent to the circle with equation \[ x^2 + y^2 = 4\]
What is the remainder of the division \[ \dfrac{x^{100} - x^{99} - x + 1}{x^2 - 3x + 2} \] .
We now need to find the values of \( a \) and \( b \) that define the remainder.
Note that \( D(x) \) may be factored as follows: \[ D(x) = x^2 - 3x + 2 = (x - 1)(x - 2) \] Hence, we have: \[ P(x) = (x - 1)(x - 2) Q(x) + ax + b \] Using the zeros of \( D(x) \), we write: \[ P(1) = (1 - 1)(1 - 2) Q(1) + a(1) + b \] This simplifies to: \[ a + b = P(1) \] Similarly, we write: \[ P(2) = (2 - 1)(2 - 2) Q(2) + a(2) + b \] This simplifies to: \[ 2a + b = P(2) \] Now we need to evaluate \( P(1) \) and \( P(2) \). \[ P(1) = 1^{100} - 1^{99} - 1 + 1 = 0 \] Next, we rewrite \( P(x) \) as: \[ P(x) = x^{99}(x - 1) - x + 1 \] Hence, we calculate \( P(2) \): \[ P(2) = 2^{99}(2 - 1) - 2 + 1 = 2^{99} - 1 \] We now have a system of equations to solve for \( a \) and \( b \): \[ a + b = 0 \] \[ 2a + b = 2^{99} - 1 \] Solving the system, we get: \[ a = 2^{99} - 1 \quad \text{and} \quad b = 1 - 2^{99} \] Thus, the remainder is: \[ r(x) = (2^{99} - 1)x + 1 - 2^{99} \]
Evaluate the number represented by the infinite series \[ \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} \]
Square both sides to obtain: \[ y^2 = \dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} \] We can write: \[ y^2 = \dfrac{1}{3} + y \] Solve the above quadratic equation to obtain: \[ y = \dfrac{3 + \sqrt{21}}{6} \quad \text{or} \quad y = \dfrac{3 - \sqrt{21}}{6} \] Since \( y \) is positive, the solution is: \[ \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} = y = \dfrac{3 + \sqrt{21}}{6} \]
Show that the 3 by 3 system of equations given below has no solutions. \[ 2x + y - 3z = 5 \] \[ -5x + 3y + 2z = 7 \] \[ 3x - 4y + z = 8 \]
Cramer's Rule gives the solution for each variable in terms of the determinants of matrices derived from the coefficient matrix \(A\) and modified by replacing columns with the constants from the right-hand side.
To find \(x\), replace the first column of \(A\) with the constants vector \(b\): \[ A_x = \begin{bmatrix} 5 & 1 & -3 \\ 7 & 3 & 2 \\ 8 & -4 & 1 \end{bmatrix} \] Then, the solution for \(x\) is: \[ x = \dfrac{\text{det}(A_x)}{\text{det}(A)} \] To find \(y\), replace the second column of \(A\) with the constants vector \(b\): \[ A_y = \begin{bmatrix} 2 & 5 & -3 \\ -5 & 7 & 2 \\ 3 & 8 & 1 \end{bmatrix} \] Then, the solution for \(y\) is: \[ y = \dfrac{\text{det}(A_y)}{\text{det}(A)} \] To find \(z\), replace the third column of \(A\) with the constants vector \(b\): \[ A_z = \begin{bmatrix} 2 & 1 & 5 \\ -5 & 3 & 7 \\ 3 & -4 & 8 \end{bmatrix} \] Then, the solution for \(z\) is: \[ z = \dfrac{\text{det}(A_z)}{\text{det}(A)} \]
Now, we need to compute the determinants of the matrices \(A\), \(A_x\), \(A_y\), and \(A_z\). \[ \text{det}(A) = \begin{vmatrix} 2 & 1 & -3 \\ -5 & 3 & 2 \\ 3 & -4 & 1 \end{vmatrix} \] Using cofactor expansion along the first row: \[ \text{det}(A) = 2 \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} - 1 \begin{vmatrix} -5 & 2 \\ 3 & 1 \end{vmatrix} + (-3) \begin{vmatrix} -5 & 3 \\ 3 & -4 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} = 3(1) - 2(-4) = 3 + 8 = 11 \] \[ \begin{vmatrix} -5 & 2 \\ 3 & 1 \end{vmatrix} = (-5)(1) - (2)(3) = -5 - 6 = -11 \] \[ \begin{vmatrix} -5 & 3 \\ 3 & -4 \end{vmatrix} = (-5)(-4) - (3)(3) = 20 - 9 = 11 \] Substitute these values into the determinant formula: \[ \text{det}(A) = 2(11) - 1(-11) + (-3)(11) = 22 + 11 - 33 = 0 \]
So, \(\text{det}(A) = 0\).
Determinant of \(A_x\): \[ \text{det}(A_x) = \begin{vmatrix} 5 & 1 & -3 \\ 7 & 3 & 2 \\ 8 & -4 & 1 \end{vmatrix} \] We can calculate this using cofactor expansion along the first row: \[ \text{det}(A_x) = 5 \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} - 1 \begin{vmatrix} 7 & 2 \\ 8 & 1 \end{vmatrix} + (-3) \begin{vmatrix} 7 & 3 \\ 8 & -4 \end{vmatrix} \] Calculate the 2x2 determinants: \[ \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} = 11, \quad \begin{vmatrix} 7 & 2 \\ 8 & 1 \end{vmatrix} = -9, \quad \begin{vmatrix} 7 & 3 \\ 8 & -4 \end{vmatrix} = -52 \] Substituting these values into the formula: \[ \text{det}(A_x) = 5(11) - 1(-9) + (-3)(-52) = 55 + 9 + 156 = 220 \] No need to continue: If one of the determnants of \(A_x\), \(A_y\) or \(A_z\) is equal to zero, then the system of equations has no solution.