Solving Equations

Equations are mathematical statements that assert the equality of two expressions. Solving equations means finding the values of variables that make the equation true. The solutions of an equation are the values of the variable that satisfy the equation, making both sides equal. For polynomial equations (where $P(x) = 0$), these solutions are also called roots or zeros. These are discussed in the separate document Finding Polynomial Roots and Zeros. In this document we will discuss (mostly) algebraic methods for solving equations of various types.

Linear Equations

Linear equations are equations of the first degree, meaning the highest power of the variable is 1.

Standard Forms

One variable: $ax + b = 0$
Two variables: $ax + by = c$
Slope-intercept form: $y = mx + b$

Solving Linear Equations

Isolate the variable by performing the same operations on both sides
Combine like terms
Use inverse operations (addition/subtraction, multiplication/division)

Example Solution Process

Example 1: Solve $2(x - 3) + 4 = 3(1 - x)$

Expand: $2x - 6 + 4 = 3 - 3x$
Simplify: $2x - 2 = 3 - 3x$
Add $3x$ to both sides: $5x - 2 = 3$
Add 2 to both sides: $5x = 5$
Divide by 5: $x = 1$

Example 2 (Combining Like Terms): Solve $3x + 7 - x + 2 = 5x - 4 + 2x$

Combine like terms on left side: $(3x - x) + (7 + 2) = 2x + 9$
Combine like terms on right side: $(5x + 2x) - 4 = 7x - 4$
Rewrite equation: $2x + 9 = 7x - 4$
Subtract $2x$ from both sides: $9 = 5x - 4$
Add 4 to both sides: $13 = 5x$
Divide by 5: $x = \frac{13}{5}$

Quadratic Equations

Quadratic equations are second-degree polynomial equations of the form: $ax^2 + bx + c = 0$ where $a \neq 0$.

Methods for Solving Quadratic Equations

1. Factoring

If the quadratic can be factored as $(px + q)(rx + s) = 0$, then:

either factor may equal 0: $px + q = 0$ or $rx + s = 0$
solve for x: $x = -\frac{q}{p}$ or $x = -\frac{s}{r}$

Example: Solve $x^2 + 7x + 12 = 0$

Find two numbers that multiply to $12$ and add to $7$: $3$ and $4$
Factor: $(x + 3)(x + 4) = 0$
Set each factor to zero: $x + 3 = 0$ or $x + 4 = 0$
Solve: $x = -3$ or $x = -4$

Note: Not all quadratics can be factored easily over the integers. For example, $x^2 + x - 1 = 0$ requires the quadratic formula.

2. Completing the Square

Completing the square is a method that rewrites the quadratic in a form that makes it easy to solve and is related to the quadratic formula (see below). The idea is to transform $ax^2 + bx + c = 0$ into $(x + d)^2 = e$ form and then use the square root property to find the roots.

Example: Solve $x^2 + 6x + 5 = 0$ by completing the square

Move constant: $x^2 + 6x = -5$
Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides: $x^2 + 6x + 9 = -5 + 9 = 4$
Factor: $(x + 3)^2 = 4$
Take square root: $x + 3 = \pm 2$
Solve: $x = -3 \pm 2$, so $x = -1$ or $x = -5$

Key difference:

Completing the square: Transform the equation step-by-step into perfect square form
Quadratic formula: Apply the pre-derived formula directly

Both give the same answer, but completing the square shows the geometric/algebraic structure more clearly.

3. Quadratic Formula

The quadratic formula is a general solution for any quadratic equation:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Using completing the square to derive the quadratic formula:

Starting with the general quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$:

Divide by $a$ (if $a \neq 1$): $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
Move constant to right side: $x^2 + \frac{b}{a}x = -\frac{c}{a}$
Take half of the coefficient of $x$ and square it: $\left(\frac{b}{2a}\right)^2$
Add this value to both sides: $x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$
Factor the left side as a perfect square: $\left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$
Take the square root of both sides: $x + \frac{b}{2a} = \pm\sqrt{-\frac{c}{a} + \left(\frac{b}{2a}\right)^2}$
Solve for $x$: $x = -\frac{b}{2a} \pm \sqrt{-\frac{c}{a} + \left(\frac{b}{2a}\right)^2}$

Example: Solve $2x^2 + 5x - 3 = 0$

Here: $a = 2$, $b = 5$, $c = -3$

\[\begin{aligned} x &= \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} \\ &= \frac{-5 \pm \sqrt{25 + 24}}{4} \\ &= \frac{-5 \pm \sqrt{49}}{4} \\ &= \frac{-5 \pm 7}{4} \end{aligned}\]

Therefore: $x = \frac{-5 + 7}{4} = \frac{1}{2}$ or $x = \frac{-5 - 7}{4} = -3$

The Discriminant:

The discriminant $\Delta = b^2 - 4ac$ determines the nature of solutions:

Two distinct real s: $\Delta > 0$
One repeated real solution: $\Delta = 0$
Two complex conjugate solutions: $\Delta < 0$

Note on Complex Conjugates: Complex conjugates are pairs of complex numbers of the form $a + bi$ and $a - bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$. They have the same real part but opposite imaginary parts.

Examples by Type:

Two real roots: $x^2 - 5x + 6 = 0$
- Discriminant: $\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0$
- Solutions: $x = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}$, so $x = 3$ or $x = 2$
One real root: $x^2 - 4x + 4 = 0$
- Discriminant: $\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0$
- Solution: $x = \frac{4 \pm \sqrt{0}}{2} = \frac{4}{2} = 2$ (double root)
Complex conjugate roots: $x^2 + x + 1 = 0$
- Discriminant: $\Delta = (1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$
- Solutions: $x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$
- Complex conjugates: $\frac{-1 + i\sqrt{3}}{2}$ and $\frac{-1 - i\sqrt{3}}{2}$

4. Graphical Method

Find x-intercepts of the parabola $y = ax^2 + bx + c$.

Steps for Graphical Solution:

Plot the parabola $y = ax^2 + bx + c$
Find x-intercepts (where the graph crosses the x-axis, i.e., where $y = 0$)
Read the x-coordinates of intersection points

Advantages:

Provides visual understanding of the solution
Shows the number of real solutions immediately
Useful for approximate solutions

Limitations:

May not give exact solutions
Requires graphing tools or software for accuracy

Example: For $y = x^2 - 3x + 2$, the parabola intersects the x-axis at $x = 1$ and $x = 2$.

Higher Degree Equations

For equations of degree 3 and higher, various specialized techniques may be needed:

Cubic Equations

Cubic equations have the form $ax^3 + bx^2 + cx + d = 0$ where $a \neq 0$.

Methods for Solving Cubic Equations

1. Factoring by Grouping

Example: Solve $x^3 + 2x^2 - x - 2 = 0$

Group terms: $(x^3 + 2x^2) + (-x - 2) = 0$
Factor each group: $x^2(x + 2) - 1(x + 2) = 0$
Factor out common factor: $(x^2 - 1)(x + 2) = 0$
Factor further: $(x - 1)(x + 1)(x + 2) = 0$
Solutions: $x = 1, -1, -2$

2. Rational Root Theorem

See 04 Finding Polynomial Roots and Zeros for details on the Rational Root Theorem and its relationship to synthetic division.

Example: Find rational roots of $2x^3 - 3x^2 - 11x + 6 = 0$

Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$
Testing $x = 3$: $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ ✓
So $(x - 3)$ is a factor.
Using synthetic division: $2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x^2 + 3x - 2)$
Factor the quadratic: $2x^2 + 3x - 2 = (2x - 1)(x + 2)$
Complete factorization: $(x - 3)(2x - 1)(x + 2) = 0$

Solutions: $x = 3, \frac{1}{2}, -2$

Substitute $u = x^2$ to get $au^2 + bu + c = 0$
Solve the quadratic in $u$ using standard methods
Back-substitute to find $x$ values from $x^2 = u$

Example

Solve: $x^4 - 5x^2 + 4 = 0$

Let $u = x^2$: $u^2 - 5u + 4 = 0$
Factor: $(u - 1)(u - 4) = 0$
Solutions: $u = 1$ or $u = 4$
Back-substitute: $x^2 = 1$ or $x^2 = 4$
Final solutions: $x = \pm 1, \pm 2$

Take logarithm of both sides: $\log(a^x) = \log(b)$
Use logarithm properties: $x \log(a) = \log(b)$
Solve: $x = \frac{\log(b)}{\log(a)}$

Example 1: Solve $2^x = 32$

Since $32 = 2^5$, we have $2^x = 2^5$, so $x = 5$.

Example 2: Solve $3^x = 50$

\[x = \frac{\log(50)}{\log(3)} = \frac{\ln(50)}{\ln(3)} \approx \frac{3.912}{1.099} \approx 3.56\]

More Complex Exponential Equations

Example: Solve $2^{x+1} + 2^x = 12$

Factor out $2^x$: $2^x \cdot 2^1 + 2^x = 12$
Simplify: $2 \cdot 2^x + 2^x = 12$
Factor: $2^x(2 + 1) = 12$
Solve: $3 \cdot 2^x = 12 \Rightarrow 2^x = 4 = 2^2$
Therefore: $x = 2$

Logarithmic Equations

Equations involving logarithmic expressions.

Basic Logarithmic Equations

Example 1: Solve $\log_2(x) = 5$

Convert to exponential form: $x = 2^5 = 32$

Example 2: Solve $\log(x) + \log(x - 3) = 1$ (common logarithm, base 10)

Use logarithm properties: $\log(x(x - 3)) = 1$
Convert to exponential form: $x(x - 3) = 10^1 = 10$
Expand: $x^2 - 3x = 10$
Rearrange: $x^2 - 3x - 10 = 0$
Factor: $(x - 5)(x + 2) = 0$
Potential solutions: $x = 5$ or $x = -2$
Check domain: $x > 0$ and $x - 3 > 0$, so $x > 3$
Valid solution: $x = 5$ (reject $x = -2$)

Equations with Multiple Logarithms

Example: Solve $\log_3(x + 1) - \log_3(x - 1) = 1$

Use quotient rule: $\log_3\left(\frac{x + 1}{x - 1}\right) = 1$
Convert to exponential form: $\frac{x + 1}{x - 1} = 3^1 = 3$
Cross multiply: $x + 1 = 3(x - 1)$
Expand: $x + 1 = 3x - 3$
Solve: $1 + 3 = 3x - x \Rightarrow 4 = 2x \Rightarrow x = 2$
Check domain: $x + 1 > 0$ and $x - 1 > 0$, so $x > 1$ ✓

Mixed Exponential-Logarithmic Equations

Example: Solve $e^x - \ln(x) = 2$

This type of equation typically requires numerical methods or graphical solutions, as there's no algebraic method to isolate $x$.

Verification and Domain Considerations

Always verify solutions by substituting back into the original equation. Pay attention to:

Domain restrictions (e.g., denominators ≠ 0, arguments of logarithms > 0)
Extraneous solutions introduced by algebraic manipulations
Practical constraints in applied problems