Other Types of Equations

Other Types of Equations

Beyond linear and quadratic equations, the ACT tests three more equation types: higher-degree polynomial equations, exponential equations, and radical equations. Each one has a single reliable method, and recognizing which method applies is most of the work.

A polynomial equation of degree three or higher is usually solved by factoring. Factoring by grouping pairs terms so that a common factor can be removed from each pair, revealing a shared binomial factor. The factor theorem connects a polynomial's zeros to its factors directly: if $x = a$ is a zero of a polynomial, then $(x - a)$ is a factor, and substituting the known zero into the polynomial and setting the result equal to zero solves for an unknown coefficient.

Factoring higher-degree polynomials

A cubic equation with no constant term factors immediately by removing the common variable factor. A cubic equation with four terms often factors by grouping the first two terms and the last two terms separately.

Worked Example. Solve by factoring: $x^3 - 4x = 0$.

Factor out the common variable factor, then factor the remaining difference of squares: $x(x^2 - 4) = 0$, so $x(x - 2)(x + 2) = 0$ and $x = 0$, $2$, or $-2$.

Never divide both sides by $x$; that throws away the solution $x = 0$. Factor it out instead.

For a four-term cubic such as $x^3 + 2x^2 - 9x - 18 = 0$, group the terms: $x^2(x + 2) - 9(x + 2) = (x + 2)(x^2 - 9) = (x + 2)(x - 3)(x + 3)$, giving $x = -2$, $3$, or $-3$. If grouping is not obvious, test small integers such as $\pm 1$, $\pm 2$, $\pm 3$ for a root, then divide it out. For $x^3 - 4x^2 - 11x + 30 = 0$, the value $x = 2$ is a root, and dividing leaves $x^2 - 2x - 15 = (x - 5)(x + 3)$.

The factor theorem

If a question says "$x = -1$ is a zero of $f(x) = x^3 + 4x^2 + kx + 2$, find $k$," substitute: $-1 + 4 - k + 2 = 0$, so $5 - k = 0$ and $k = 5$. Take care with negative numbers raised to odd powers; $(-1)^3 = -1$, not $1$.

Exponential equations

Rewriting both sides of an exponential equation with the same base converts the equation into a simpler statement about exponents alone, since two powers of the same base are equal only when their exponents are equal. This requires no logarithms at this level.

Worked Example. Solve $9^x = 27^{x-1}$.

Rewrite both sides with base $3$: $(3^2)^x = (3^3)^{x-1}$, so $3^{2x} = 3^{3x-3}$. Set the exponents equal: $2x = 3x - 3$, so $x = 3$.

Useful powers to know by heart: $2^1$ through $2^6$ ($2, 4, 8, 16, 32, 64$), $3^1$ through $3^4$ ($3, 9, 27, 81$), and $5^1$ through $5^3$ ($5, 25, 125$). Negative exponents turn a fraction into a power: $\frac{1}{16} = 2^{-4}$. When the exponent is itself an expression, as in $(3^3)^{x-1}$, multiply the exponents, distributing across the whole expression: $3(x - 1) = 3x - 3$. Forgetting to distribute is the most common error. Also watch the final question: if it asks for $x^2 + 1$, solve for $x$ first and then compute $x^2 + 1$.

Exponential models appear in decay problems. A medication amount modeled by $A(t) = 200(0.5)^{t/6}$ drops to $25$ milligrams after $18$ hours, found by writing $25$ as $200$ times a power of one-half ($25 = 200 \cdot (0.5)^3$) and matching exponents: $\frac{t}{6} = 3$.

Radical equations

A radical equation is solved by isolating the radical and squaring both sides. Squaring both sides of an equation can introduce a solution that satisfies the squared equation but not the original equation, so every candidate solution must be checked in the original equation before being accepted.

Worked Example. Solve $\sqrt{3x + 1} = x - 1$.

Squaring gives $3x + 1 = x^2 - 2x + 1$, so $x^2 - 5x = 0$ and $x = 0$ or $x = 5$. Check $x = 0$: $\sqrt{1} = 1$, but $0 - 1 = -1$, so $x = 0$ is extraneous. Check $x = 5$: $\sqrt{16} = 4$ and $5 - 1 = 4$, which matches. The only solution is $x = 5$.

A quick warning sign: when the right side of $\sqrt{\ \ } = (\text{expression})$ turns out negative for a candidate, that candidate fails, because a square root is never negative.

Strategy checklist

• Cubic with no constant term: factor out $x$ first. Four terms: try grouping.
• Unknown coefficient and a known zero: substitute the zero and set the result equal to $0$.
• Exponential equation: same base, then set exponents equal, distributing carefully.
• Radical equation: isolate, square, solve, and check every answer in the original equation.
• Finish by answering the quantity asked for, not just $x$.