Carefully selected Algebra 2 problems designed to challenge high school students and strengthen their mathematical reasoning are presented. Each problem is accompanied by a clear and detailed solution to support understanding and mastery.
Solve the quadratic equation: \[ 3x^2 - 5x - 2 = 0 \]
Use the quadratic formula: \[ x =\dfrac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} \] \[ = \dfrac{5 \pm \sqrt{25 + 24}}{6} = \dfrac{5 \pm \sqrt{49}}{6} \] \[ x = \dfrac{5 \pm 7}{6} \Rightarrow x = \dfrac{12}{6} = 2 \quad \text{or} \quad x = \dfrac{-2}{6} = -\dfrac{1}{3} \]
Simplify the expression: \[ \dfrac{x^2 - 9}{x^2 - x - 6} \]
Factor numerator and denominator: \[ \dfrac{x^2 - 9}{x^2 - x - 6} = \dfrac{(x - 3)(x + 3)}{(x - 3)(x + 2)} \] Cancel the common term \( (x - 3) \) \[ \dfrac{x^2 - 9}{x^2 - x - 6} = \dfrac{x + 3}{x + 2}, \quad x \ne 3 \]
Solve the system of equations: \[ \begin{cases} 2x + y = 7 \\ x^2 + y^2 = 25 \end{cases} \]
From the first equation: \[ y = 7 - 2x \] Substitute into the second equation: \[ x^2 + (7 - 2x)^2 = 25 \Rightarrow x^2 + 49 - 28x + 4x^2 = 25 \Rightarrow 5x^2 - 28x + 24 = 0 \] Solve with the quadratic formula: \[ x = \dfrac{28 \pm \sqrt{784 - 480}}{10} = \dfrac{28 \pm \sqrt{304}}{10} \] \[ x = \dfrac{28 \pm 4\sqrt{19}}{10} = \dfrac{14 \pm 2\sqrt{19}}{5} \] Find \( y \) by substituting \( x \) by its value in \( y = 7 - 2x \) \[ y = 7 - 2 \left( \dfrac{14 \pm 2\sqrt{19}}{5}\right) \] Simplify \[ \begin{pmatrix}x=\dfrac{2\left(7-\sqrt{19}\right)}{5},\:&y=\dfrac{7+4\sqrt{19}}{5}\\ x=\dfrac{2\left(7+\sqrt{19}\right)}{5},\:&y=\dfrac{7-4\sqrt{19}}{5}\end{pmatrix} \]
Find the inverse \( f^{-1}(x) \) of the function: \[ f(x) = \dfrac{2x - 1}{x + 3} \]
Let \( y = \dfrac{2x - 1}{x + 3} \) and interchange \( x \) and \( y \): \[ x = \dfrac{2y - 1}{y + 3} \] Cross multiply: \[ x(y + 3) = 2y - 1 \] Expand the left side: \[ xy + 3x = 2y - 1 \] Group all terms with \( y \) on one side \[xy - 2y = -3x - 1\] Factor \( y \). \[ y(x - 2) = -3x - 1 \] Divide both sides by \( (x - 2) \) and simplfy to obtain: \[ y = \dfrac{-3x - 1}{x - 2} \] The inverse function is given by: \[ f^{-1}(x) = \dfrac{-3x - 1}{x - 2} \]
If \( f(x) = x^2 + 2x + 3 \) and \( g(x) = \sqrt{x + 4} \), find the domain of \( (g \circ f)(x) \).
By definition: \[ (g \circ f)(x) = g(f(x)) = \sqrt{f(x) + 4} \] The term \( f(x) + 4 \) inside the square root must be non negative, hence: \[ f(x) + 4 \geq 0 \] Substitute \( f(x) \) by its expression: \[ x^2 + 2x + 3 + 4 \geq 0 \] Group \[ x^2 + 2x + 7 \geq 0 \] The discriminant of the quadratic expression \( x^2 + 2x + 7 \) is: \[ \Delta = b^2 - 4 a c = 2^2 - 4(1)(7) = - 24 \] Since the discriminant is negative and the leading coefficient of the quadratic expression \( a = 1 \) is positive, this quadratic expression is always positive and therefore the ineqiuality \( x^2 + 2x + 3 + 4 \geq 0 \) is satisfied for all real values of \( x \) , so the domain of \( (g \circ f)(x) \) is: \[ (-\infty, +\infty) \]
Solve the inequality: \[ \dfrac{x + 2}{x - 3} > 1 \]
Subtract \( 1 \) from both sides and simplify: \[ \dfrac{x + 2}{x - 3} - 1 > 0 \] Write \( 1 \) as \( \dfrac{x - 3}{x - 3} \): \[ \dfrac{x + 2}{x - 3} - \dfrac{x - 3}{x - 3} > 0 \] Subtract the fractional expressions as they have common denominator: \[ \dfrac{x + 2 - (x - 3)}{x - 3} > 0 \] \[ \dfrac{5}{x - 3} > 0 \] This inequality is satisfied for\[ x - 3 > 0 \Rightarrow x > 3 \] Using interval notation, the solution set is written as: \[ (3 , + \infty) \]
A geometric sequence has first term \( a = 5 \) and common ratio \( r = 3 \). What is the 6th term?
\[ a_n = a \cdot r^{n - 1} = 5 \cdot 3^{5} = 5 \cdot 243 = 1215 \]
Given \( \log_2(x + 3) + \log_2(x - 1) = 3 \), solve for \( x \).
First find the domain of values of \( x \) where the solution must be. The arguments of the logarithmic expressions must be positive, hence the conditions: \[ x + 3 \gt 0 \quad \text{and} \quad x - 1 > 0 \] The solution set to the above inequalities, which is the domain of the equation, is: \[ (1 , +\infty ) \] Use product rule: \[ \log_2[(x + 3)(x - 1)] = 3 \] Transform the logarithmic equation to exponential equations: \[ (x + 3)(x - 1) = 2^3 = 8 \] Expand the left side and group like terms: \[ x^2 + 2x - 3 = 8 \Rightarrow x^2 + 2x - 11 = 0 \] Use quadratic formulas to solve the above quadratic equation: \[ x = \dfrac{-2 \pm \sqrt{48}}{2} = \dfrac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3} \] Check for domain and only \[ x = - 1 + 2 \sqrt 3 \] is a solution to the given equation.
Let \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \), find \( A^{-1} \) if it exists.
The formula for the inverse of a 2 by 2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by: \[ A^{-1} = \dfrac{1}{\det A} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] For \[ A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \] Determinant: \[ \det A = (2)(4) - (-1)(3) = 8 + 3 = 11 \] Inverse: \[ A^{-1} = \dfrac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix} \]
Write the equation of a parabola with vertex at \( (2, -3) \) and passing through the point \( (4, 5) \).
Use vertex form: \[ y = a(x - 2)^2 - 3 \] Substitute point: \[ 5 = a(4 - 2)^2 - 3 \Rightarrow 5 = 4a - 3 \Rightarrow a = 2 \] Equation: \[ y = 2(x - 2)^2 - 3 \]
Given that one root of the cubic polynomial \( f(x) = x^3 - 6x^2 + 11x - k \) is 2, find all the roots of \( f(x) \).
Use synthetic division to divide by \( x - 2 \): \[ \begin{array}{r|rrrr} 2 & 1 & -6 & 11 & -k \\ & & 2 & -8 & 6 \\ \hline & 1 & -4 & 3 & 6 - k \end{array} \] The remainder is \( 6 - k \).
For 2 to be a root, the remainder of the synthetic division \( 6 - k \) must be euqal to 0: \[ 6 - k = 0 \Rightarrow k = 6 \] Substitute \( k \) by \( 6 \) in the division above: \[ \begin{array}{r|rrrr} 2 & 1 & -6 & 11 & - 6 \\ & & 2 & -8 & 6 \\ \hline & 1 & -4 & 3 & 0 \end{array} \] Using the synthetic division, we write: \[ f(x) = (x - 2)(x^2 - 4x + 3) \] Factor \( x^2 - 4x + 3 \) \[ f(x) = (x - 2)(x - 1)(x - 3) \] So the roots of \( f(x) \) are \( 1, 2, 3 \).
If \( f(x) = 3^{x+1} - 2 \) and \( f(a) = f(b) \), prove that \( a = b \). Then explain what this tells you about the function.
If \( f(a) = f(b) \), then: \[ 3^{a+1} - 2 = 3^{b+1} - 2 \Rightarrow 3^{a+1} = 3^{b+1} \] Since exponential functions are one-to-one: \[ a + 1 = b + 1 \Rightarrow a = b \] Conclusion: \( f(x) \) is one-to-one and has an inverse function.
Solve for all real values of \( x \): \[ \left| \dfrac{2x - 5}{x + 1} \right| = 3 \]
Domain of given equation: \( x \ne -1 \)
Split into two cases:
Case 1: \[ \dfrac{2x - 5}{x + 1} = 3 \] Cross multiply and solve: \[ 2x - 5 = 3(x + 1) \Rightarrow 2x - 5 = 3x + 3\] \[ \Rightarrow -8 = x \Rightarrow x = -8 \] Case 2: \[ \dfrac{2x - 5}{x + 1} = -3 \] Cross multiply and solve: \[2x - 5 = -3(x + 1) \Rightarrow 2x - 5 = -3x - 3 \Rightarrow 5x \] \[ = 2 \Rightarrow x = \dfrac{2}{5} \] Check solutions: For \( x = - 8 \) \[ \left| \dfrac{2x - 5}{x + 1} \right| = \left| \dfrac{2(-8) - 5}{-8 + 1} \right| = 3 \] Hence \( x = - 8 \) is a solution of the given equation.
For \( x = \dfrac{2}{5} \) \[ \left| \dfrac{2x - 5}{x + 1} \right| = \left| \dfrac{2 \left( \dfrac{2}{5} \right) - 5}{ \dfrac{2}{5} + 1} \right| = 3 \] Hence \( x = \dfrac{2}{5} \) is also a solution of the given equation.
The given equation has two solutions: \[ \left\{ -8 , \dfrac{2}{5} \right\} \]
Let \( f(x) = x^4 - 10x^2 + 9 \). Factor \( f(x) \) completely over the real numbers.
Let \( y = x^2 \) and hence \( y^2 = x^4 \): \[ x^4 - 10x^2 + 9 = y^2 - 10y + 9 = (y - 1)(y - 9) \] Now substitute back: \[ f(x) = x^4 - 10x^2 + 9 = (x^2 - 1)(x^2 - 9) \] \[ = (x - 1)(x + 1)(x - 3)(x + 3) \]
Find all real values of \( x \) that satisfy the equation: \[ \sqrt{2x + 3} - \sqrt{-x + 7} = 1 \]
Isolate each radical on opposite sides. Add \(\sqrt{-x + 7}\) to both sides and simplify: \[ \sqrt{2x + 3} = \sqrt{-x + 7} + 1 \] Square both sides to eliminate the first square root \[ \left(\sqrt{2x + 3}\right)^2 = \left(\sqrt{-x + 7} + 1\right)^2 \] \[ 2x + 3 = (-x + 7) + 2\sqrt{-x + 7} + 1 \] \[ 2x + 3 = -x + 8 + 2\sqrt{-x + 7} \] Isolate the remaining square root. Subtract \(-x + 8\) from both sides and simplify: \[ 2x + 3 + x - 8 = 2\sqrt{-x + 7} \] \[ 3x - 5 = 2\sqrt{-x + 7} \] Square both sides again \[ (3x - 5)^2 = (2\sqrt{-x + 7})^2 \] \[ 9x^2 - 30x + 25 = 4(-x + 7) \] \[ 9x^2 - 30x + 25 = -4x + 28 \] Bring all terms to one side \[ 9x^2 - 30x + 25 + 4x - 28 = 0 \] \[ 9x^2 - 26x - 3 = 0 \] Use the quadratic formula: \[ x = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(9)(-3)}}{2(9)} \] \[ x = \frac{26 \pm \sqrt{676 + 108}}{18} \] \[ x = \frac{26 \pm \sqrt{784}}{18} \] \[ x = \frac{26 \pm 28}{18} \] \[ x = \frac{54}{18} = 3 \quad \text{or} \quad x = \frac{-2}{18} = -\frac{1}{9} \] Check both solutions in the original equation
Check \(x = 3\): \[ \sqrt{2(3) + 3} - \sqrt{-3 + 7} = \sqrt{6 + 3} - \sqrt{4} = \sqrt{9} - 2 = 3 - 2 = 1 \quad \checkmark \] Check \(x = -\frac{1}{9}\): \[ \sqrt{2(-\frac{1}{9}) + 3} - \sqrt{-(-\frac{1}{9}) + 7} = \sqrt{-\frac{2}{9} + 3} - \sqrt{\frac{1}{9} + 7} \] \[ = \sqrt{\frac{25}{9}} - \sqrt{\frac{64}{9}} = \frac{5}{3} - \frac{8}{3} = -1 \neq 1 \quad \text{Reject} \] The real solution to the given equation is: \[ x = 3 \]
Let \( f(x) = \dfrac{x^2 - 4x + 3}{x^2 - 5x + 6} \). Determine all vertical and horizontal asymptotes, and describe the end behavior.
Factor both: \[ f(x) = \dfrac{(x - 1)(x - 3)}{(x - 2)(x - 3)} \Rightarrow f(x) = \dfrac{x - 1}{x - 2}, \quad x \ne 3 \] There is a Hole at: \[ x = 3 \] Vertical asymptote: \[ f(x) = \dfrac{x - 1}{x - 2}, \quad x \ne 3 \] Set denominator equal to zero. \[ x - 2 = 0 \] Solve to find vertical asymptote: \[ x = 2 \]
Horizontal asymptote: degrees of the numerator and the denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficient in the numerator and denominator. \[ \text{ratio of leading coefficient: } \dfrac{1}{1} = 1 \Rightarrow y = 1 \] Horizontal asymptote given by: \[ y = 1 \]
End behavior: As \( x \to \pm\infty \), \( f(x) \to 1 \) meaning that the graph stays close to the vertical asymptote.
Solve for all real values of \( x \): \[ \log_3(x^2 - 4x) = \log_3(3x - 8) \]
Since logarithms have the same base and are equal, their arguments are equal: \[ x^2 - 4x = 3x - 8 \] Rewrite the equation as: \[ x^2 - 7x + 8 = 0 \] Use quadratic formulas: \[ x = \dfrac{7 \pm \sqrt{49 - 32}}{2} = \dfrac{7 \pm \sqrt{17}}{2} \] Because when solving, we did not take into account the domain of the given equation, we now need to check the solutions of the given equation by calculating the left hand side (LHS) and right hand side (RHS) of the equation and compere them..
1) For \[ x = \dfrac{7 + \sqrt{17}}{2} \] \[ \text{LHS} = \log_3(x^2 - 4x) = \log_3((\dfrac{7 + \sqrt{17}}{2})^2 - 4(\dfrac{7 + \sqrt{17}}{2})) \] \[ = \log _3\left(5+3\sqrt{17}\right)-\log _3\left(2\right) \] \[ \text{RHS} = \log_3(3x - 8) = \log_3(3(\dfrac{7 + \sqrt{17}}{2}) - 8) \] \[ = \log _3\left(5+3\sqrt{17}\right)-\log _3\left(2\right) \] Conclusion: LHS = RHS and therefore \( x = \dfrac{7 + \sqrt{17}}{2} \) is a solution to the given equation.
2) For \[ x = \dfrac{7 - \sqrt{17}}{2} \] \[ \text{LHS} = \log_3(x^2 - 4x) = \log_3((\dfrac{7 - \sqrt{17}}{2})^2 - 4(\dfrac{7 - \sqrt{17}}{2})) \] \[ \approx \log_3( -3.68465 ) \] The LHS is not real because the argument of the log is negative.
No need to check the RHS for \( x = \dfrac{7 - \sqrt{17}}{2} \) because the LHS is not real.
Conclusion: The solution of the given equation is: \[ x = \dfrac{7 + \sqrt{17}}{2} \]
The sum of the first \( n \) terms of a geometric sequence is \( S_n = 81 \left(1 - \left(\dfrac{1}{3}\right)^n\right) \). Find the smallest \( n \) for which \( S_n > 80.99 \).
\[ S_n > 80.99 \Rightarrow 81 \left(1 - \left(\dfrac{1}{3}\right)^n\right) > 80.99 \] \[ \Rightarrow 1 - \left(\dfrac{1}{3}\right)^n > \dfrac{80.99}{81} \] \[ \Rightarrow - \left(\dfrac{1}{3}\right)^n > \dfrac{80.99}{81} - 1 \] Mutliply both sides of the inequality by \( -1 \) and change symbol of inequality \[ \Rightarrow \left(\dfrac{1}{3}\right)^n < 1 - \dfrac{80.99}{81} \] Take log of both sides of the inequality: \[ n \log\left(\dfrac{1}{3}\right) < \log(\frac{1}{8100}) \] \[ \Rightarrow n > \dfrac{\log(\frac{1}{8100})}{\log(1/3)} \approx \dfrac{-3.908}{-0.4771} \approx 8.2 \] Since \( n \) is a positive integer, its smallest value so that \( S_n > 80.99 \) is the next integer greater than \( 8.2 \): \[ n = 9 \]
Let \( f(x) = \sqrt{x + 2} \) and \( g(x) = \dfrac{1}{x - 1} \). Find the domain of \( (f \circ g)(x) \).
By definition: \[ (f \circ g)(x) = f(g(x)) \] \[ f(g(x)) = \sqrt{\dfrac{1}{x - 1} + 2} \Rightarrow \dfrac{1}{x - 1} + 2 \geq 0 \] \[ \Rightarrow \dfrac{1 + 2(x - 1)}{x - 1} \geq 0 \Rightarrow \dfrac{2x - 1}{x - 1} \geq 0 \] Solve using sign chart:
Undefined at \( x = 1 \)
Critical points: \( x = \dfrac{1}{2}, x = 1 \)
Test intervals for \( \dfrac{2x - 1}{x - 1} \):
For \( x < \dfrac{1}{2} \): \( \quad \dfrac{2x - 1}{x - 1} \) is positive
For \( \dfrac{1}{2} < x < 1 \): \( \quad \dfrac{2x - 1}{x - 1} \) is negative
For \( x > 1 \): \( \quad \dfrac{2x - 1}{x - 1} \) is positive
The domain of \( (f \circ g)(x) \) is given by: \[ (-\infty , \dfrac{1}{2} ) \cup (1, \infty) \]
Let \( A = \begin{bmatrix} x & 2 \\ 1 & x \end{bmatrix} \). Find all values of \( x \) for which \( A \) is not invertible.
A square matrix is non-invertible when its determinant is equal to zero: \[ \det(A) = x^2 - 2 = 0 \] The equation \( x^2 - 2 = 0 \) has two solurions : \( x = \sqrt 2 \) and \( x = - \sqrt 2 \)
Conclusion: Matrix \( A \) is invertible over real all numbers except \( x = \pm \sqrt 2 \).