1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations

Quadratic, Polynomial, Exponential & Logarithmic Equations

Non-linear equations appear frequently in the ACET Mathematics subtest. Mastering polynomial root behaviors, quadratic optimizations, exponential growth curves, and logarithmic properties will allow you to solve complex equations quickly.

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1. Quadratic Function & Equation Analysis

General Form: $ax^2 + bx + c = 0 \quad (a \neq 0)$

The Quadratic Formula & The Discriminant ($\Delta$)

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\Delta = b^2 - 4ac$$

• $\Delta > 0$ and a perfect square: Two distinct, rational real roots.
• $\Delta > 0$ and non-square: Two distinct, irrational real roots.
• $\Delta = 0$: One real repeated root (double root, parabola tangent to x-axis).
• $\Delta < 0$: Two complex conjugate non-real roots (parabola does not touch x-axis).

Vieta's Formulas (Quadratic & Higher Degree)

For $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$:

$$\text{Sum of Roots: } r_1 + r_2 = -\frac{b}{a}$$ $$\text{Product of Roots: } r_1 r_2 = \frac{c}{a}$$ $$\text{Sum of Reciprocals: } \frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2} = \frac{-b/a}{c/a} = -\frac{b}{c}$$ $$\text{Sum of Squares: } r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2 - 2ac}{a^2}$$

Vertex Form & Optimization

Vertex Form: $y = a(x - h)^2 + k$

• Vertex coordinates: $h = -\frac{b}{2a}$, $k = c - \frac{b^2}{4a} = f(h)$.
• If $a > 0$, vertex is a minimum at $y = k$.
• If $a < 0$, vertex is a maximum at $y = k$.

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2. Polynomial Theorems & Root Analysis

For a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$:

1. Remainder Theorem: Dividing $P(x)$ by $(x - c)$ leaves remainder $R = P(c)$.
2. Factor Theorem: $(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$.
3. Rational Root Theorem: Any rational root $\frac{p}{q}$ (in lowest terms) must have $p$ as a factor of constant term $a_0$ and $q$ as a factor of leading coefficient $a_n$.
4. Descartes' Rule of Signs: The number of positive real roots equals the number of sign variations in $P(x)$ or is less by an even integer.

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3. Exponential Equations & Applications

An exponential function has the form $f(x) = a \cdot b^x \quad (b > 0, b \neq 1)$.

Compound Interest & Continuous Growth Models

• Periodic Compounding: $A = P \left(1 + \frac{r}{n}\right)^{nt}$
• Continuous Compounding: $A = P e^{rt}$
• Radioactive Decay / Half-Life: $N(t) = N_0 \left(\frac{1}{2}\right)^{t / t_{1/2}}$

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4. Logarithmic Properties & Equations

Definition: $y = \log_b(x) \iff b^y = x \quad (x > 0, b > 0, b \neq 1)$.

Core Logarithmic Laws

1. Product Rule: $\log_b(x y) = \log_b(x) + \log_b(y)$
2. Quotient Rule: $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$
3. Power Rule: $\log_b(x^k) = k \log_b(x)$
4. Change of Base Formula: $\log_b(a) = \frac{\log_c(a)}{\log_c(b)} = \frac{\ln a}{\ln b}$
5. Inverse Identities: $b^{\log_b(x)} = x$ and $\log_b(b^x) = x$

CRITICAL ACET TRAP: $\log_b(x + y) \neq \log_b(x) + \log_b(y)$. Logarithms do NOT distribute across addition.

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Worked Step-by-Step ACET Exam Problems

Problem 1: Root Manipulations Using Vieta's Formulas

Without explicitly solving the equation $2x^2 - 8x + 5 = 0$, find the exact numerical value of $r_1^2 + r_2^2$, where $r_1$ and $r_2$ are the roots of the equation.

Step 1: Identify coefficients. $a = 2$, $b = -8$, $c = 5$.

Step 2: Apply Vieta's formulas for sum and product.

$$r_1 + r_2 = -\frac{-8}{2} = 4$$ $$r_1 r_2 = \frac{5}{2} = 2.5$$

Step 3: Express $r_1^2 + r_2^2$ using identity.

$$r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2$$ $$r_1^2 + r_2^2 = (4)^2 - 2(2.5) = 16 - 5 = 11$$

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Problem 2: Solving an Exponential Equation via Common Bases

Solve for $x$: $9^{x + 1} - 28 \cdot 3^x + 3 = 0$.

Step 1: Express terms with a common base of 3. Notice $9^{x + 1} = (3^2)^{x + 1} = 3^{2x + 2} = 3^2 \cdot (3^x)^2 = 9 (3^x)^2$.

$$9 (3^x)^2 - 28 (3^x) + 3 = 0$$

Step 2: Substitute $u = 3^x$ (where $u > 0$).

$$9 u^2 - 28 u + 3 = 0$$

Step 3: Factor the quadratic in $u$.

$$(9u - 1)(u - 3) = 0 \implies u = \frac{1}{9} \quad \text{or} \quad u = 3$$

Step 4: Solve back for $x$.

• If $3^x = \frac{1}{9} = 3^{-2} \implies x = -2$.
• If $3^x = 3^1 \implies x = 1$.

Both solutions $x = -2$ and $x = 1$ are valid.

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ACET Speed Tactics & Traps Summary