Algebra Problems and Questions with Solutions - Grade 12

We rewrite the given expressions to the same power so that we can easily compare their bases. To do this, we use the power of a power rule: \( (x^a)^b = x^{a \cdot b} \).

Let's rewrite all expressions to have an outer exponent of 100:

\[ 25^{100} \text{ (Already has an exponent of 100)} \]

\[ 2^{300} = (2^3)^{100} = 8^{100} \]

\[ 3^{400} = (3^4)^{100} = 81^{100} \]

\[ 4^{200} = (4^2)^{100} = 16^{100} \]

\[ 2^{600} = (2^6)^{100} = 64^{100} \]

Now that the exponents are identical, we order them from greatest to least by simply comparing the base numbers (81, 64, 25, 16, 8):

\[ \mathbf{3^{400} , \quad 2^{600}, \quad 25^{100} , \quad 4^{200} , \quad 2^{300}} \]

According to the Rational Root Theorem, the possible rational zeros of a polynomial \( P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \) are given by:

\[ \dfrac{\text{factors of the constant term } a_0}{\text{factors of the leading coefficient } a_n} \]

Identify the leading coefficient and the constant term from the given polynomial:

The leading coefficient is \( 1 \), and its factors are: \( \pm 1 \)

The constant term is \( 6 \), and its factors are: \( \pm 1, \pm 2, \pm 3, \pm 6 \)

Therefore, dividing the factors of the constant by the factors of the leading coefficient gives the possible rational zeros:

\[ \pm 1, \pm 2, \pm 3, \pm 6 \]

Testing for Rational Zeros, using direct substitution and calculations to see which ones equal 0:

\[ P(1) = (1)^3 - 7(1) + 6 = 1 - 7 + 6 = 0 \]

\[ P(2) = (2)^3 - 7(2) + 6 = 8 - 14 + 6 = 0 \]

\[ P(-3) = (-3)^3 - 7(-3) + 6 = -27 + 21 + 6 = 0 \]

Since these values satisfy \( P(x) = 0 \), they are the zeros of \( P(x) \). The polynomial \( P(x) \) has the following rational zeros:

\[ \mathbf{x = 1, \quad x = 2, \quad x = -3} \]

Thus, the polynomial can be factored completely using these roots in the form \( (x - \text{root}) \):

\[ \mathbf{P(x) = (x-1)(x-2)(x+3)} \]

We analyze the x-intercepts (zeros) on the graph to determine the factors and their multiplicities. If the graph touches the axis and turns around, it has an even multiplicity. If it crosses straight through, it has an odd multiplicity.

• From the graph, at \( x = -3 \), the curve touches the x-axis and turns around. This means \( x = -3 \) is a zero of multiplicity 2. This gives us the factor \( (x + 3)^2 \).
• At \( x = 0 \), the curve crosses straight through the axis. This means \( x = 0 \) is a zero of multiplicity 1. This gives us the factor \( x \).
• At \( x = 2 \), the curve touches the x-axis and turns around again. This means \( x = 2 \) is a zero of multiplicity 2. This gives us the factor \( (x - 2)^2 \).

Adding the multiplicities (\( 2 + 1 + 2 \)), the polynomial has a lowest possible degree of 5.

Next, we determine the sign of the leading coefficient by looking at the end behavior. The graph is falling on the right and rising on the left. An odd-degree polynomial with this end behavior must have a negative leading coefficient.

We are told the absolute value of the leading coefficient is 1, so the leading coefficient must be exactly \( -1 \).

Combining the leading coefficient and the factors, \( P(x) \) is given by:

\[ \mathbf{P(x) = - x (x + 3)^2 (x - 2)^2} \]

According to the Complex Conjugate Root Theorem, since the coefficients of the polynomial \( P(x) \) are real numbers, any complex roots must occur in conjugate pairs. Thus, if \( 2 - i \) is a zero, then \( 2 + i \) is also a zero of \( P(x) \).

Next, we find the quadratic factor corresponding to these two complex roots by multiplying them together:

\[ (x - (2 - i))(x - (2 + i)) \]

Expanding this expression is easier if we regroup it to form a difference of squares:

\[ = (x - 2 + i)(x - 2 - i) \]

\[ = (x - 2)^2 - i^2 \]

Recall that the imaginary unit squared is \( -1 \) (i.e., \( i^2 = -1 \)). Substitute this into the equation:

\[ = (x - 2)^2 - (-1) = (x - 2)^2 + 1 \]

Now expand the perfect square:

\[ = x^2 - 4x + 4 + 1 = x^2 - 4x + 5 \]

Since \( P(x) \) is divisible by this quadratic factor \( x^2 - 4x + 5 \), we perform polynomial long division to find the remaining factors:

\[ (x^4 - 4x^3 + 3x^2 + 8x - 10) \div (x^2 - 4x + 5) \]

Dividing, we obtain a remainder of 0 and the quotient \( x^2 - 2 \). This quotient can be factored further as a difference of squares:

\[ x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2}) \]

Setting these factors to zero gives us the remaining roots. Thus, the complete set of zeros of \( P(x) \) is:

\[ \mathbf{2 - i, \quad 2 + i, \quad -\sqrt{2}, \quad \sqrt{2}} \]

We use the vertex form of a quadratic function, which makes it easy to plug in the coordinates of the vertex directly:

\[ f(x) = a(x - h)^2 + k \]

Substitute the given vertex \( (h, k) = (-2,1) \) into the equation:

\[ f(x) = a (x + 2)^2 + 1 \]

Now, use the given point \( (0, -3) \) to solve for the unknown multiplier \( a \). Substitute \( x = 0 \) and \( f(0) = -3 \):

\[ -3 = a(0 + 2)^2 + 1 \]

Simplify the squared term:

\[ -3 = 4a + 1 \]

Subtract 1 from both sides and solve for \( a \):

\[ 4a = -4 \implies a = -1 \]

Substituting \( a = -1 \) back into the vertex form yields our complete equation:

\[ f(x) = -(x + 2)^2 + 1 \]

To find the values of \( a \), \( b \), and \( c \), we must expand this equation into the standard form \( ax^2 + bx + c \):

\[ f(x) = -(x^2 + 4x + 4) + 1 \]

Distribute the negative sign:

\[ f(x) = -x^2 - 4x - 4 + 1 \]

Combine the constant terms:

\[ f(x) = -x^2 - 4x - 3 \]

Comparing this with the standard quadratic form \( f(x) = ax^2 + bx + c \), we find:

\[ \mathbf{a = -1, \quad b = -4, \quad c = -3} \]

Method 1: Algebraic Substitution

The standard form of a quadratic function is:

\[ f(x) = ax^2 + bx + c \]

Given \( f(1) = 3 \), we substitute \( x = 1 \) to create our first equation:

\[ 3 = a(1)^2 + b(1) + c \implies 3 = a + b + c \quad \text{(Equation B)} \]

Given \( f(5) = 3 \), we substitute \( x = 5 \) to create our second equation:

\[ 3 = a(5)^2 + b(5) + c \implies 3 = 25a + 5b + c \quad \text{(Equation C)} \]

Subtracting Equation B from Equation C allows us to eliminate the variable \( c \):

\[ (25a + 5b + c) - (a + b + c) = 3 - 3 \]

\[ 24a + 4b = 0 \]

Divide by 4 to simplify: \( 6a + b = 0 \), which implies \( b = -6a \).

The x-coordinate of the vertex for any quadratic is given by the formula \( \dfrac{-b}{2a} \). Substituting \( -6a \) in for \( b \):

\[ \dfrac{-(-6a)}{2a} = \dfrac{6a}{2a} = \mathbf{3} \]

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Note that this problem might be solved much faster using geometry as follows:

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. Because they have the exact same y-coordinate (3), they represent horizontally symmetrical points on the graph.

Since the function is quadratic, the x-coordinate of the vertex is located exactly halfway between any two symmetrical points. Therefore, it is equal to the x-coordinate of the midpoint of these two points.

Calculate the midpoint's x-coordinate:

\[ x_{\text{vertex}} = \dfrac{1 + 5}{2} = \dfrac{6}{2} = \mathbf{3} \]

The \( x \)-coordinate of the vertex of the graph of \( f(x) \) is 3.

The oblique (or slant) asymptote of a rational function is the quotient resulting from the polynomial long division of the numerator by the denominator.

Let's perform long division of \( a x^4 + b x^3 + 3 \) by \( x^3 - 2 \):

1. Divide the leading term \( ax^4 \) by \( x^3 \), which gives \( ax \). This is the first term of the quotient.

2. Multiply \( ax \) by the entire denominator \( (x^3 - 2) \), which yields \( ax^4 - 2ax \).

3. Subtract this from the numerator: \( (ax^4 + bx^3 + 3) - (ax^4 - 2ax) = bx^3 + 2ax + 3 \).

4. Now, divide the new leading term \( bx^3 \) by \( x^3 \), which gives \( b \). This is the second term of the quotient.

Using long division, we obtain the full expression:

\[ \dfrac{ax^4 + bx^3 + 3}{x^3 - 2} = ax + b + \dfrac{2ax + (3 + 2b)}{x^3 - 2} \]

The quotient \( ax + b \) is the oblique asymptote. We are given that the oblique asymptote must be equal to the line \( 2x - 3 \).

Therefore, we set them equal to each other:

\[ ax + b = 2x - 3 \]

Since two polynomials are equal if and only if their corresponding coefficients are equal, we compare the terms and obtain:

\[ \mathbf{a = 2, \quad b = -3} \]

Given the equation:

\[ \log_9 (x^3) = \log_2(8) \]

First, simplify the right-hand side of the given equation. Since 8 can be written as \( 2^3 \), the logarithm simplifies easily:

\[ \log_2(8) = \log_2 (2^3) = 3 \]

Rewrite the original equation substituting this new value:

\[ \log_9 (x^3) = 3 \]

To solve for \( x \), we can rewrite the number 3 using a logarithm with base 9 so that both sides match, using the rule \( c = \log_b(b^c) \):

\[ \log_9 (x^3) = \log_9(9^3) \]

Since the logarithmic bases on both sides are identical, we can drop the logs and write the algebraic equation for the arguments:

\[ x^3 = 9^3 \]

Take the cube root of both sides to solve for \( x \):

\[ \mathbf{x = 9} \]

Given the initial logarithmic equation:

\[ \log_x(y^3) = 2 \]

To establish an algebraic relationship between \( x \) and \( y \), rewrite the equation from logarithmic form into exponential form (\( \text{base}^{\text{answer}} = \text{argument} \)):

\[ x^2 = y^3 \]

The target expression we want to evaluate is \( \log_y(x^4) \), which contains an \( x^4 \). We need to manipulate our exponential equation to create an \( x^4 \). Square both sides of the equation:

\[ (x^2)^2 = (y^3)^2 \]

Simplify using the power of a power rule:

\[ x^4 = y^6 \]

Now, substitute this value for \( x^4 \) back into the target expression we want to evaluate. Take the log of base \( y \) of both sides and simplify:

\[ \log_y(x^4) = \log_y(y^6) \]

Using the property of logarithms \( \log_b(b^k) = k \), the expression evaluates to:

\[ \mathbf{6} \]

Given the logarithmic equation:

\[ \log_x (8e^3) = 3 \]

To solve for the base \( x \), rewrite the equation from logarithmic form into exponential form (\( \text{base}^{\text{answer}} = \text{argument} \)):

\[ x^3 = 8e^3 \]

To isolate \( x \), we reverse the cube power by taking the cube root on both sides of the equation:

\[ x = \sqrt[3]{8e^3} \]

We can simplify the expression inside the radical. Since \( 8 \) is a perfect cube (\( 8 = 2^3 \)), we substitute it and group the terms:

\[ x = \sqrt[3]{2^3 e^3} = \sqrt[3]{(2 e)^3} \]

The cube root and the cube power cancel each other out, leaving the final answer:

\[ \mathbf{x = 2e} \]

Given the exponential equation:

\[ 16^x + 16^{x - 1} = 10 \]

First, use the exponential rules to separate and rewrite the terms so they share a common base of 4. We know \( 16 = 4^2 \):

\[ 16^x = (4^2)^x = 4^{2 x} \]

For the second term, use the rule \( a^{m-n} = \frac{a^m}{a^n} \) and the substitution from above:

\[ 16^{x - 1} = 16^x \cdot 16^{-1} = \dfrac{4^{2x}}{16} \]

Substitute these simplified terms back into the given equation:

\[ 4^{2x} + \dfrac{4^{2x}}{16} = 10 \]

Factor out the common term \( 4^{2x} \) to isolate it. \( 4^{2x}(1 + \frac{1}{16}) = 10 \), which becomes \( 4^{2x}(\frac{17}{16}) = 10 \). Multiply both sides by \( \frac{16}{17} \) to solve for \( 4^{2x} \):

\[ 4^{2x} = \dfrac{160}{17} \]

To reduce the base from 4 to 2, take the positive square root of both sides of the above equation (exponential functions are always positive):

\[ \sqrt {4^{2x}} = \sqrt {(4^x)^2} = 4^x = \dfrac{\sqrt{160}}{\sqrt{17}} = \dfrac{4\sqrt{10}}{\sqrt{17}} \]

Hence, since \( 4^x \) is mathematically identical to \( 2^{2x} \) (because \( 4^x = (2^2)^x = 2^{2x} \)), we have our answer:

\[ \mathbf{2^{2x} = \dfrac{4\sqrt{10}}{\sqrt{17}}} \]

Note: This question could also be solved starting by converting directly to base 2.

Substitute \( 16^x = (2^4)^x = 2^{4 x} \) directly into the equation and follow similar factoring steps as in the given solution above.

Given the first equation:

\[ a^2 - b^2 = 8 \]

Because we want to find the sum of terms raised to the power of 4, we must square both sides of the given equation to generate those higher powers:

\[ (a^2 - b^2)^2 = 8^2 \]

Expand the left side using the perfect square binomial rule \( (x-y)^2 = x^2 - 2xy + y^2 \):

\[ a^4 - 2a^2b^2 + b^4 = 64 \]

Rearranging the terms, we get an expression containing our desired target \( a^4 + b^4 \):

\[ a^4 + b^4 - 2a^2b^2 = 64 \quad \text{(I)} \]

Now we use the second piece of given information:

\[ ab = 2 \]

Square both sides of the above equation so it matches the middle term of our expanded expression:

\[ (ab)^2 = a^2b^2 = 2^2 = 4 \]

Substitute this value of \( 4 \) in place of \( a^2b^2 \) in equation (I):

\[ a^4 + b^4 - 2(4) = 64 \]

Multiply to simplify:

\[ a^4 + b^4 - 8 = 64 \]

Add 8 to both sides to find the final value. Hence:

\[ \mathbf{a^4 + b^4 = 72} \]

To find the minimum and maximum values of a complex trigonometric function, we build it from the inside out. We know the fundamental range of a basic sine function is always between -1 and 1, regardless of its internal arguments:

\[ -1 \leq \sin\left( 2x - \dfrac{\pi}{3} \right) \leq 1 \]

Multiply all terms of the double inequality by the amplitude coefficient 2:

\[ -2 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) \leq 2 \]

Add the vertical shift of -5 to all terms of the inequality:

\[ -2 - 5 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \leq 2 - 5 \]

Simplify the bounds:

\[ -7 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \leq -3 \]

Now, change the above inequality by applying the absolute value operation. Because the entire expression inside the absolute value bars ranges strictly in the negative numbers (from -7 to -3), taking the absolute value flips the signs to positive and reverses the numerical order of the bounds:

\[ |-3| \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| \leq |-7| \]

\[ 3 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| \leq 7 \]

Finally, add the outside constant 3 to all terms of the double inequality to complete the function \( f(x) \):

\[ 3 + 3 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| + 3 \leq 7 + 3 \]

Simplify the bounds to reveal the final range:

\[ 6 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| + 3 \leq 10 \]

Therefore, the maximum value of \( f(x) \) is equal to 10 and the minimum value of \( f(x) \) is equal to 6.

When simplifying nested absolute values, we work from the inside out. We use the given domain condition to determine the sign of the expression inside the bars. If the expression is negative, we multiply it by -1 to evaluate the absolute value.

We are given the condition: If \( x \lt -7 \), then by adding 4 to both sides, we know that:

\[ x \lt -3 \]

Add 3 to the above inequality and simplify to determine the sign of the inner expression \( (3 + x) \):

\[ x + 3 \lt 0 \]

Since \( (3+x) \) is definitively less than zero, according to the definition of the absolute value, we must multiply it by -1 to remove the bars:

\[ |3 + x| = -(3 + x) \]

Substitute this back in. The given expression simplifies to:

\[ |4 - |3 + x|| = |4 - (- (3 + x))| = |4 + 3 + x| = |x + 7| \]

Now we evaluate the outer absolute value \( |x + 7| \). We return to our original condition \( x \lt -7 \). Adding 7 to both sides reveals that:

\[ x + 7 \lt 0 \]

Since the new expression is also definitively negative, we again multiply by -1 to evaluate the absolute value:

\[ |x + 7| = - (x + 7) = - x - 7 \]

And hence, the fully simplified expression is:

\[ \mathbf{|4 - |3 + x|| = -x - 7} \]

Average speed is not calculated by simply averaging two speeds together; it must be calculated using the formula: Total Distance divided by Total Time. We must set up expressions for distance and time.

Let \( d \) be the distance between \( A \) and \( B \). The time taken to travel the first leg from A to B is calculated by dividing distance by speed:

\[ t_1 = \dfrac{d}{50} \]

where 50 km/hour is the average speed from \( A \) to \( B \).

Let the unknown 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 round trip is the sum of both times:

\[ t_{\text{total}} = t_1 + t_2 = \dfrac{d}{50} + \dfrac{d}{v} \]

The total distance covered (away and return) is simply twice the distance:

\[ d + d = 2d \]

The average speed for the whole trip is given by the total distance over total time. Set this formula equal to our target average speed of 60 km/hour:

\[ \text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{2d}{t_{\text{total}}} \]

Substitute the expressions we built into the formula:

\[ \dfrac{2d}{\dfrac{d}{50} + \dfrac{d}{v}} = 60 \]

Notice that the distance \( d \) is present in every term. Cancel \( d \) from both the numerator and denominator by dividing it out:

\[ \dfrac{2}{\dfrac{1}{50} + \dfrac{1}{v}} = 60 \]

To clear the fraction, multiply both sides by the denominator expression \( \left( \dfrac{1}{50} + \dfrac{1}{v} \right) \):

\[ 2 = 60 \left( \dfrac{1}{50} + \dfrac{1}{v} \right) \]

Distribute the 60 to simplify the terms inside the parentheses:

\[ 2 = \dfrac{60}{50} + \dfrac{60}{v} \]

\[ 2 = 1.2 + \dfrac{60}{v} \]

Subtract 1.2 from both sides to isolate the term with the variable:

\[ 0.8 = \dfrac{60}{v} \]

Solve for \( v \) by multiplying by \( v \) and dividing by 0.8:

\[ v = \dfrac{60}{0.8} = 75 \]

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.

Given the non-linear equation, we can use the algebraic identity for a difference of squares to factor it into simpler binomials:

\[ x^2 - y^2 = (x - y)(x + y) = -12 \]

We are given a second equation: \( x + y = 6 \). Substitute the expression \( (x + y) \) in our factored equation with the number 6:

\[ (x - y)(6) = -12 \]

Divide both sides by 6 to solve for \( x - y \):

\[ (x - y) = -2 \]

We now have a simple linear system of equations to solve using elimination or substitution:

\[ \begin{cases} x - y = -2 \\ x + y = 6 \end{cases} \]

Add the two equations together vertically to eliminate the \( y \) variable: \( (x+x) + (y-y) = -2 + 6 \), which gives \( 2x = 4 \), meaning \( x = 2 \).

Substitute \( x = 2 \) back into the second equation: \( 2 + y = 6 \), which gives \( y = 4 \).

Solve the system above gives the final answer:

\[ \mathbf{x = 2, \quad y = 4} \]

Given the functional equation:

\[ f(x) + 3f(8 - x) = x \]

To evaluate a specific value like \( f(2) \), we need to substitute strategic values for \( x \) to build a solvable system of equations. First, substitute \( x = 2 \) into the given equation:

\[ f(2) + 3 f(8-2) = 2 \]

\[ f(2) + 3 f(6) = 2 \tag{A} \]

This creates an equation with a new unknown, \( f(6) \). To find a way to eliminate it, substitute \( x = 6 \) into the original given equation:

\[ f(6) + 3 f(8-6) = 6 \]

\[ f(6) + 3 f(2) = 6 \tag{B} \]

We now have a system of two equations. We can solve equation (B) to isolate \( f(6) \):

\[ f(6) = 6 - 3f(2) \tag{C} \]

Substitute this expression for \( f(6) \) back into equation (A):

\[ f(2) + 3(6 - 3f(2)) = 2 \]

Expand the expression by distributing the 3 and group the like terms:

\[ f(2) + 18 - 9f(2) = 2 \]

\[ - 8 f(2) + 18 = 2 \]

Subtract 18 from both sides and solve for \( f(2) \):

\[ -8 f(2) = -16 \]

\[ \mathbf{f(2) = 2} \]

Given the recursive functional equation:

\[ f(2x + 1) = 2 f(x) + 1 \tag{A} \]

We must use the known value \( f(0) = 2 \) as a stepping stone to find \( f(3) \). We need to determine what values of \( x \) will produce the function evaluations we need.

To find the next step in the chain from \( x=0 \), let \( x = 0 \) in equation \( A \):

\[ f(2(0) + 1) = 2 f(0) + 1 \]

Substitute the known value \( f(0) = 2 \) into the equation:

\[ f(1) = 2(2) + 1 = 4 + 1 = 5 \]

Now we know \( f(1) = 5 \). We want to find \( f(3) \). We need the inner argument of the left side, \( (2x+1) \), to equal 3. This occurs when \( x = 1 \). Let \( x = 1 \) in equation \( A \):

\[ f(2(1) + 1) = 2 f(1) + 1 \]

\[ f(3) = 2 f(1) + 1 \tag{B} \]

Substitute our newly calculated value \( f(1) = 5 \) into equation \( B \) to obtain the final answer:

\[ f(3) = 2(5) + 1 \]

\[ \mathbf{f(3) = 11} \]

Let us find the points of intersections of the circle and the line by solving the system of the two given equations. A line that is tangent to a circle intersects it at exactly one point, meaning the resulting system should have exactly one solution.

Substitute the expression for \( y \), which is \( 2 x + b \), into the equation of the circle:

\[ x^2 + (2 x + b)^2 = 4 \]

Expand the squared binomial and simplify the equation by combining like terms:

\[ x^2 + 4x^2 + 4bx + b^2 = 4 \]

\[ 5 x^2 + 4 b x + (b^2 - 4) = 0 \]

The number of points of intersection is given by the number of solutions of the above quadratic equation. The line and circle are tangent if the above quadratic equation has only one solution. A quadratic equation has exactly one solution when its discriminant (\( \Delta = B^2 - 4AC \)) is equal to zero.

Find the discriminant \( \Delta \) as a function of \( b \), using \( A=5 \), \( B=4b \), and \( C=(b^2 - 4) \):

\[ \Delta = ( 4 b )^2 - 4 (5)(b^2 - 4) \]

Set the discriminant equal to zero and solve for \( b \):

\[ ( 4 b )^2 - 4 (5)(b^2 - 4) = 0 \]

\[ 16b^2 - 20b^2 + 80 = 0 \]

\[ -4b^2 + 80 = 0 \]

\[ -4b^2 = -80 \implies b^2 = 20 \]

Take the square root of both sides. Solve for \( b \) to obtain two solutions:

\[ b = \pm \sqrt{20} = \pm \sqrt{4 \cdot 5} \]

\[ \mathbf{b = \pm 2 \sqrt{5}} \]

Let the numerator polynomial be \( P(x) = x^{100} - x^{99} - x + 1 \) and the divisor polynomial be \( D(x) = x^2 - 3x + 2 \).

The division of the two polynomials may be written algebraically as:

\[ P(x) = D(x) Q(x) + r(x) \]

where \( Q(x) \) is the quotient and \( r(x) \) is the remainder. By definition of polynomial division, the remainder will have a degree equal to one or lower because the divisor \( D(x) \) has a degree of 2. Thus, we can express the remainder as a linear equation \( r(x) = ax + b \).

We now need to find the values of \( a \) and \( b \) that define the remainder.

Note that \( D(x) \) may be factored into binomials as follows:

\[ D(x) = x^2 - 3x + 2 = (x - 1)(x - 2) \]

Hence, we substitute the factored divisor and linear remainder into our division equation:

\[ P(x) = (x - 1)(x - 2) Q(x) + ax + b \]

Using the zeros of \( D(x) \) (which are 1 and 2), we can evaluate \( P(x) \). Because substituting a zero of the divisor will make the entire quotient term multiply by zero, it isolates our remainder terms. First, we write the equation for \( x=1 \):

\[ P(1) = (1 - 1)(1 - 2) Q(1) + a(1) + b \]

This simplifies to:

\[ a + b = P(1) \]

Similarly, we write the equation for \( x=2 \):

\[ P(2) = (2 - 1)(2 - 2) Q(2) + a(2) + b \]

This simplifies to:

\[ 2a + b = P(2) \]

Now we need to evaluate \( P(1) \) and \( P(2) \) using the original polynomial equation.

Substitute \( x=1 \) into \( P(x) \):

\[ P(1) = 1^{100} - 1^{99} - 1 + 1 = 1 - 1 - 1 + 1 = 0 \]

Next, we rewrite \( P(x) \) by factoring it slightly to make evaluating larger numbers easier:

\[ P(x) = x^{99}(x - 1) - (x - 1) \]

Hence, we calculate \( P(2) \) by substituting \( x=2 \):

\[ P(2) = 2^{99}(2 - 1) - (2 - 1) = 2^{99}(1) - 1 = 2^{99} - 1 \]

We now have a linear system of equations to solve for \( a \) and \( b \):

\[ \begin{cases} a + b = 0 \\ 2a + b = 2^{99} - 1 \end{cases} \]

From the first equation, we know \( b = -a \). Substitute this into the second equation: \( 2a - a = 2^{99} - 1 \implies a = 2^{99} - 1 \).

Solving the system, we get:

\[ a = 2^{99} - 1 \quad \text{and} \quad b = -(2^{99} - 1) = 1 - 2^{99} \]

Thus, substituting \( a \) and \( b \) back into \( r(x) = ax+b \), the remainder is:

\[ \mathbf{r(x) = (2^{99} - 1)x + 1 - 2^{99}} \]

We can solve this nested radical by setting the entire infinite expression equal to a variable.

Let \( y = \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} \).

Square both sides of the equation to eliminate the outermost square root:

\[ y^2 = \dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} \]

Because the series of square roots is infinite, the expression underneath the first radical is identical to our original definition of the variable \( y \). Therefore, we can substitute \( y \) into the right side of the equation. We can write:

\[ y^2 = \dfrac{1}{3} + y \]

This is a standard quadratic equation. Rearrange it to set it equal to zero so we can solve it. Multiply by 3 to remove fractions, yielding \( 3y^2 = 1 + 3y \), which becomes:

\[ 3y^2 - 3y - 1 = 0 \]

Use the quadratic formula (\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)) to solve the above quadratic equation to obtain:

\[ y = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(3)(-1)}}{2(3)} = \dfrac{3 \pm \sqrt{9 + 12}}{6} \]

\[ y = \dfrac{3 + \sqrt{21}}{6} \quad \text{or} \quad y = \dfrac{3 - \sqrt{21}}{6} \]

Since the original expression is a principal square root, \( y \) must evaluate to a positive number. The second option yields a negative result because \( \sqrt{21} > 3 \). Since \( y \) is positive, the solution is the positive root:

\[ \mathbf{\sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} = y = \dfrac{3 + \sqrt{21}}{6}} \]

We can determine the solvability of a system by extracting its matrices and evaluating their determinants. The coefficient matrix \(A\), composed of the variables' multipliers, is:

\[ A = \begin{bmatrix} 2 & 1 & -3 \\ -5 & 3 & 2 \\ 3 & -4 & 1 \end{bmatrix} \]

The constants matrix \(b\), composed of the values on the right side of the equals sign, is:

\[ b = \begin{bmatrix} 5 \\ 7 \\ 8 \end{bmatrix} \]

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. If the determinant of \(A\) is zero, but a modified determinant is not, the system has no solution.

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 given by the fraction: \( x = \dfrac{\text{det}(A_x)}{\text{det}(A)} \)

To find \(y\), replace the second column of \(A\) with the constants vector \(b\):

\[ A_y = \begin{bmatrix} 2 & 5 & -3 \\ -5 & 7 & 2 \\ 3 & 8 & 1 \end{bmatrix} \implies y = \dfrac{\text{det}(A_y)}{\text{det}(A)} \]

To find \(z\), replace the third column of \(A\) with the constants vector \(b\):

\[ A_z = \begin{bmatrix} 2 & 1 & 5 \\ -5 & 3 & 7 \\ 3 & -4 & 8 \end{bmatrix} \implies z = \dfrac{\text{det}(A_z)}{\text{det}(A)} \]

Now, we need to compute the determinants of the matrices \(A\) and \(A_x\). First, evaluate matrix \(A\):

\[ \text{det}(A) = \begin{vmatrix} 2 & 1 & -3 \\ -5 & 3 & 2 \\ 3 & -4 & 1 \end{vmatrix} \]

Using cofactor expansion along the first row (multiplying each top-row element by the determinant of the 2x2 matrix that remains when crossing out its row and column):

\[ \text{det}(A) = 2 \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} - 1 \begin{vmatrix} -5 & 2 \\ 3 & 1 \end{vmatrix} + (-3) \begin{vmatrix} -5 & 3 \\ 3 & -4 \end{vmatrix} \]

Calculating the 2x2 determinants using the formula \( (ad - bc) \):

\[ \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} = 3(1) - 2(-4) = 3 + 8 = 11 \]

\[ \begin{vmatrix} -5 & 2 \\ 3 & 1 \end{vmatrix} = (-5)(1) - (2)(3) = -5 - 6 = -11 \]

\[ \begin{vmatrix} -5 & 3 \\ 3 & -4 \end{vmatrix} = (-5)(-4) - (3)(3) = 20 - 9 = 11 \]

Substitute these calculated values back into the main determinant formula:

\[ \text{det}(A) = 2(11) - 1(-11) + (-3)(11) = 22 + 11 - 33 = 0 \]

So, we have established that \(\text{det}(A) = 0\). This means the system does not have a unique solution. Now we check the determinant of the modified matrix \(A_x\):

\[ \text{det}(A_x) = \begin{vmatrix} 5 & 1 & -3 \\ 7 & 3 & 2 \\ 8 & -4 & 1 \end{vmatrix} \]

We calculate this using cofactor expansion along the first row in the exact same manner:

\[ \text{det}(A_x) = 5 \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} - 1 \begin{vmatrix} 7 & 2 \\ 8 & 1 \end{vmatrix} + (-3) \begin{vmatrix} 7 & 3 \\ 8 & -4 \end{vmatrix} \]

Calculate the 2x2 determinants:

\[ \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} = 11, \quad \begin{vmatrix} 7 & 2 \\ 8 & 1 \end{vmatrix} = 7 - 16 = -9, \quad \begin{vmatrix} 7 & 3 \\ 8 & -4 \end{vmatrix} = -28 - 24 = -52 \]

Substituting these values into the expansion formula:

\[ \text{det}(A_x) = 5(11) - 1(-9) + (-3)(-52) = 55 + 9 + 156 = 220 \]

There is no need to continue calculating \(A_y\) or \(A_z\). Because Cramer's Rule requires division by \( \text{det}(A) \), evaluating \( x \) would result in \( \dfrac{220}{0} \), which is undefined. If one of the determinants of \(A_x\), \(A_y\) or \(A_z\) is non-zero while the main determinant is zero, the system of equations is inconsistent and has no solution.

To combine and add logarithms, they must share the exact same base. We can use the change of base formula, \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), to convert all logarithms in the equation to base 2.

First, convert the base-4 term. We know \( \log_2(4) = 2 \):

\[ \log_4(x) = \dfrac{\log_2(x)}{\log_2(4)} = \dfrac{\log_2(x)}{2} = \dfrac{1}{2}\log_2(x) \]

Next, convert the base-16 term. We know \( \log_2(16) = 4 \):

\[ \log_{16}(x) = \dfrac{\log_2(x)}{\log_2(16)} = \dfrac{\log_2(x)}{4} = \dfrac{1}{4}\log_2(x) \]

Now substitute these converted terms back into the original equation:

\[ \log_2(x) + \dfrac{1}{2}\log_2(x) + \dfrac{1}{4}\log_2(x) = 7 \]

Since all terms now share the \( \log_2(x) \) variable, we can factor it out:

\[ \log_2(x) \left(1 + \dfrac{1}{2} + \dfrac{1}{4}\right) = 7 \]

Find a common denominator (which is 4) to add the fractions inside the parentheses:

\[ \log_2(x) \left(\dfrac{4}{4} + \dfrac{2}{4} + \dfrac{1}{4}\right) = 7 \]

\[ \dfrac{7}{4}\log_2(x) = 7 \]

Multiply both sides by the reciprocal \( \frac{4}{7} \) to isolate the logarithm:

\[ \log_2(x) = 4 \]

Finally, convert the equation from logarithmic form to exponential form (\( \text{base}^{\text{answer}} = \text{argument} \)) to solve for \( x \):

\[ x = 2^4 \]

\[ \mathbf{x = 16} \]

This system looks intimidating, but we can use the standard algebraic identity for the sum of two cubes to factor the first equation and simplify the system:

\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]

Substitute the known values from the original equations into this identity. We know the cubes sum to 35, and the base variables sum to 5:

\[ 35 = (5)(x^2 - xy + y^2) \]

Divide both sides by 5 to isolate the quadratic expression:

\[ x^2 - xy + y^2 = 7 \quad \text{(Equation A)} \]

We need another equation to compare this to. Let's look at the expression \( (x+y)^2 \). If we take the second given equation \( (x+y = 5) \) and square both sides, we get:

\[ (x+y)^2 = 5^2 \]

Expand the perfect square binomial:

\[ x^2 + 2xy + y^2 = 25 \quad \text{(Equation B)} \]

We can now subtract Equation A from Equation B to eliminate the squared terms and isolate \( xy \):

\[ (x^2 + 2xy + y^2) - (x^2 - xy + y^2) = 25 - 7 \]

\[ 3xy = 18 \]

Divide by 3:

\[ xy = 6 \]

We have simplified the complex cubic system into a much simpler puzzle: we are looking for two numbers whose sum is 5 (\( x+y=5 \)) and whose product is 6 (\( xy=6 \)). Geometrically, these are the roots of a quadratic equation in the form \( t^2 - (\text{sum})t + (\text{product}) = 0 \).

\[ t^2 - 5t + 6 = 0 \]

Factoring the quadratic yields \( (t-2)(t-3) = 0 \), meaning the two numbers must be 2 and 3.

Since the original system of equations is perfectly symmetric for \( x \) and \( y \), either variable can take either value. The solutions are the coordinate pairs:

\[ \mathbf{(2, 3) \quad \text{and} \quad (3, 2)} \]

To solve a rational inequality, we must first find all critical points by factoring both the numerator and the denominator completely.

Both the top and bottom expressions are differences of squares. Let's factor them:

\[ \dfrac{(x - 3)(x + 3)}{(x - 2)(x + 2)} \leq 0 \]

The critical points occur where the numerator is zero (representing the x-intercepts) and where the denominator is zero (representing vertical asymptotes where the function is undefined).

• Numerator zeros: \( x = 3, x = -3 \). Because the inequality is "less than or equal to" (\( \leq \)), the function is allowed to equal zero. Therefore, these points will be included in our final solution (using solid brackets).
• Denominator zeros: \( x = 2, x = -2 \). Because division by zero is mathematically undefined, the function can never exist at these points. Therefore, these points will never be included in our final solution (using parentheses).

We place these four critical points on a number line in numerical order: -3, -2, 2, 3. This splits the number line into five distinct intervals. We must test a random value within each interval to see if the overall fractional expression evaluates to a positive or negative number:

• Interval \( (-\infty, -3) \): Test \( x = -4 \). The signs of the factored terms are \( \frac{(-)(-)}{(-)(-)} \). A negative divided by a negative is positive.
• Interval \( (-3, -2) \): Test \( x = -2.5 \). The signs are \( \frac{(-)(+)}{(-)(-)} \). A negative divided by a positive is negative.
• Interval \( (-2, 2) \): Test \( x = 0 \). The signs are \( \frac{(-)(+)}{(-)(+)} \). A negative divided by a negative is positive.
• Interval \( (2, 3) \): Test \( x = 2.5 \). The signs are \( \frac{(-)(+)}{(+)(+)} \). A negative divided by a positive is negative.
• Interval \( (3, \infty) \): Test \( x = 4 \). The signs are \( \frac{(+)(+)}{(+)(+)} \). A positive divided by a positive is positive.

The inequality asks us to find where the expression is less than or equal to 0, meaning we want the Negative intervals. These are the intervals between -3 and -2, and between 2 and 3.

Applying our inclusion rules (brackets for numerator roots, parentheses for undefined denominator roots), we write the final interval notation:

\[ \mathbf{[-3, -2) \cup (2, 3]} \]