Algebra Questions for Grade 11 with Solutions

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 perfect 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 $-\dfrac{9}{2} + 4$:

$$ = 2\left(x - \dfrac{3}{2}\right)^2 - \dfrac{1}{2} $$

The coordinates of the points of intersection of the parabola and the line are found by solving the system of equations:

$$ \begin{cases} y = x^2 - 5x + 4 \\ y = 2x - 2 \end{cases} $$

Substitute $y$ with $2x - 2$ in the first equation:

$$ 2x - 2 = x^2 - 5x + 4 $$

Write the quadratic equation in standard form (one side equal to zero):

$$ x^2 - 7x + 6 = 0 $$

Solution of the quadratic equation:

$$ x = 1 \quad \text{and} \quad x = 6 $$

Find the $y$ coordinates:

For $x = 1$, $y = 2(1) - 2 = 0$

For $x = 6$, $y = 2(6) - 2 = 10$

Points of intersection:

$$ (1, 0) \quad \text{and} \quad (6, 10) $$

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 $k$:

$$ k = -13 $$

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:

$$ \text{Center is at: } (1, -2), \quad \text{Radius = } 4 $$

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 strictly positive:

$$ 25 + 8k > 0 $$

Solve for $k$:

$$ k > -\frac{25}{8} $$

According to Cramer's rule, if the determinant $D$ of the coefficient matrix is equal to zero and one of the determinants $D_x$ or $D_y$ is not equal to zero, the system has no solutions.

The coefficient matrix $A$ is:

$$ A = \begin{bmatrix} 2 & k \\ 5 & -3 \end{bmatrix} $$

Its determinant $D$ is:

$$ D = \begin{vmatrix} 2 & k \\ 5 & -3 \end{vmatrix} = (2)(-3) - (5)(k) = -6 - 5k $$

Setting $D = 0$:

$$ -6 - 5k = 0 \quad \Rightarrow \quad k = -\frac{6}{5} $$

We define determinant $D_x$ by replacing the first column of $A$ with the constants matrix $\begin{bmatrix} 2 \\ 7 \end{bmatrix}$:

$$ D_x = \begin{vmatrix} 2 & k \\ 7 & -3 \end{vmatrix} = (2)(-3) - (7)(k) = -6 - 7k $$

Substituting $k = -\frac{6}{5}$:

$$ D_x = -6 - 7\left(-\frac{6}{5}\right) = -6 + \frac{42}{5} = \frac{-30 + 42}{5} = \frac{12}{5} \neq 0 $$

Since the coefficient determinant $D = 0$ but $D_x \neq 0$, the system is inconsistent and has no solutions when:

$$ k = -\frac{6}{5} $$

In the quadratic expression $ax^2 + bx + c$, we identify $a = 6$, $b = -13$, and $c = 5$. Multiply $a$ and $c$:

$$ 6 \times 5 = 30 $$

We need two numbers that multiply to 30 and add to -13. The numbers -10 and -3 satisfy these conditions:

$$ -10 \times -3 = 30 $$

$$ -10 + (-3) = -13 $$

We use these numbers to split $-13x$ as $-10x - 3x$:

$$ 6x^2 - 10x - 3x + 5 $$

Group the terms and factor out common elements:

$$ (6x^2 - 10x) + (-3x + 5) = 2x(3x - 5) - 1(3x - 5) $$

Since $(3x - 5)$ is common, the expression is factored as:

$$ 6x^2 - 13x + 5 = (2x - 1)(3x - 5) $$

Note that $i^4 = 1$. Also, observe that $231 = 4 \times 57 + 3$. Hence:

$$ i^{231} = i^{4 \times 57 + 3} = i^{4 \times 57} \cdot i^3 = (i^4)^{57} \cdot i^3 $$

Since $i^4 = 1$, we get:

$$ = 1^{57} \cdot (-i) = -i $$

According to the remainder theorem, the remainder $r$ of the division of $f(x)$ by $x - c$ is given by $f(c)$. Here, $c = 1$.

$$ r = f(1) = (1 - 2)^{54} = (-1)^{54} = 1 $$

The formula for the x-coordinate of the vertex $h$ is $-\frac{b}{2a}$:

$$ h = \frac{- (-b)}{2(4)} = \frac{b}{8} = 2 $$

Solve for $b$:

$$ b = 16 $$

The vertex point is a solution to the equation of the parabola, so substitute $x=2$ and $y=4$:

$$ 4(2)^2 - 16(2) - c = 4 $$

$$ 16 - 32 - c = 4 \quad \Rightarrow \quad -16 - c = 4 $$

Solve for $c$:

$$ c = -20 $$

Hence $b = 16$ and $c = -20$.

Since $x = 2$ is a root, $(x - 2)$ is a factor of $P(x)$. Divide $P(x)$ by $(x - 2)$ to find the quadratic quotient $Q(x)$:

$$ Q(x) = \dfrac{x^3 - 3x^2 - 10x + 24}{x - 2} = x^2 - x - 12 $$

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:

$$ P(x) = (x - 2)(x - 4)(x + 3) $$

Set each factor to zero to list all the roots:

$$ x = 2, \quad x = 4, \quad x = -3 $$

Solve the compound inequality by subtracting 2 from all parts:

$$ 3 \lt 2x \lt 7 $$

Divide by 2:

$$ \frac{3}{2} \lt x \lt \frac{7}{2} $$

Since $\frac{7}{2} = 3.5$, the greatest integer value of $x$ which satisfies the given inequality is $3$ (the largest integer strictly less than $3.5$).

a) The only common element to both $A$ and $B$ is 3, hence the intersection is:

$$ A \cap B = \{3\} $$

b) All elements of $A$ and $B$ are in the union. Elements common to both are listed only once:

$$ A \cup B = \{2, 3, 5, 6, 7, 8, 9, 10\} $$

Rewrite the expression by factoring out the negative sign:

$$ = | - (x^2 - 4x + 4) | $$

Complete the square:

$$ = | -(x - 2)^2 | $$

Use the property that $|-A| = |A|$. Since $(x-2)^2$ is always non-negative for any real number $x$, we can drop the absolute value bars entirely:

$$ = (x - 2)^2 $$

A line and a circle are tangent if they have exactly one point of intersection. Substitute $y = kx$ into the circle equation:

$$ (x - 3)^2 + (kx - 5)^2 = 4 $$

Expand:

$$ x^2 - 6x + 9 + k^2x^2 - 10kx + 25 = 4 $$

Group the terms to write the quadratic equation in standard form $(Ax^2 + Bx + C = 0)$:

$$ (1 + k^2)x^2 - (6 + 10k)x + 30 = 0 $$

For the line to be tangent, this quadratic must have exactly one solution, meaning its discriminant $\Delta$ must equal zero:

$$ \Delta = B^2 - 4AC = (-(6 + 10k))^2 - 4(1 + k^2)(30) = 0 $$

Expand the discriminant equation:

$$ (36 + 120k + 100k^2) - 120 - 120k^2 = 0 $$

$$ -20k^2 + 120k - 84 = 0 $$

Divide the entire equation by $-4$ for easier solving:

$$ 5k^2 - 30k + 21 = 0 $$

Solve this quadratic equation using the quadratic formula for $k$:

$$ k = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(5)(21)}}{2(5)} = \frac{30 \pm \sqrt{900 - 420}}{10} = \frac{30 \pm \sqrt{480}}{10} $$

Simplify the radical ($\sqrt{480} = 4\sqrt{30}$):

$$ k = \frac{30 \pm 4\sqrt{30}}{10} = \frac{15 \pm 2\sqrt{30}}{5} $$

Final Answers:

$$ k = \frac{15+2\sqrt{30}}{5} \quad \text{or} \quad k = \frac{15-2\sqrt{30}}{5} $$