Grade 11 Math: Exponents, Equations & Quadratic Patterns

Start by noticing that $8 = 2^3$, so we can rewrite the equation as $2^{x+1} = (2^3)^x$. This becomes $2^{x+1} = 2^{3x}$, so $x+1=3x$, so $2x=1$, so $x=\frac{1}{2}$. Alternatively, we may write the equation as $2^{x+1} = 2^{3x}$ and since the bases are equal, the exponents are equal, so $x+1=3x$, so $2x=1$, so $x=\frac{1}{2}$. So, the solution is $x=\frac{1}{2}$.

Start by factorising the left-hand side of the inequality as $(x-3)(x+1) > 0$. The roots are $x = 3$ and $x = -1$, and the graph of the corresponding quadratic curve is above the $x$-axis for $x < -1$ and $x > 3. So, the solution is $x < -1$ or $x > 3.

Since the quadratic expression has equal roots, we know the discriminant is zero. Therefore, $(5)^2 - 4(1)(6) = 0$, so $25 - 24 = 0$, which is true, so it has equal roots. To find $x$, we use the quadratic formula: $x = \frac{-5 \pm \sqrt{0}}{2} = -\frac{5}{2}$. So, the solution is $x = -\frac{5}{2}.

We can substitute $n = 1, 2, 3$ into the general term $an^2 + bn + c$ to obtain a system of three equations in three unknowns: $a + b + c = 5$, $4a + 2b + c = 12$, and $9a + 3b + c = 21. Solving the system yields $a = 2, $b = -2$, and $c = 5. So, the solution is $a = 2, $b = -2, and $c = 5.

Using the definition of $f(x), we can calculate $f(x+1) = (x+1)^2 - 4(x+1) + 3 = x^2 + 2x + 1 - 4x - 4 + 3 = x^2 - 2x + 2 = f(x) - 2x + 1. So, the solution is $f(x+1) = f(x) - 2x + 1.