Complete Number Sets, Algebra, & Equations
The ultimate, comprehensive guide to IMAT mathematical foundations. We explore the complete hierarchy of number sets, advanced factorization techniques, exhaustive equation solving, and the rigorous logic required for inequalities.
Part 1: The Ultimate Hierarchy of Numbers
Before manipulating numbers through algebraic operations, we must precisely define what species of numbers we are dealing with. Mathematics categorizes numbers into distinct sets, each expanding upon the limitations of the previous one. Understanding these definitions is strictly required for logical reasoning questions.
- Natural Numbers ($\mathbb{N}$): The most basic numbers used for counting items. Typically defined as $\{1, 2, 3, 4, ...\}$. Note that some mathematical conventions include zero in this set, but strictly speaking, the counting numbers start at 1.
- Integers ($\mathbb{Z}$): We introduce zero and the concept of negative quantities (debt, temperature below zero). The set is $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.
- Rational Numbers ($\mathbb{Q}$): The word rational stems from "ratio". Any number that can be expressed as a perfect ratio of two integers $\frac{p}{q}$ (where $q \neq 0$) is rational. This powerful set includes all integers (since $5 = \frac{5}{1}$), terminating decimals ($0.25 = \frac{1}{4}$), and infinitely repeating decimals.
- Irrational Numbers: Numbers that absolutely cannot be expressed as a simple fraction. Their decimal expansions are both non-terminating and non-repeating. Classic examples include $\sqrt{2}, \sqrt{3}, \pi$, and Euler's number $e$.
- Real Numbers ($\mathbb{R}$): The complete union of Rational and Irrational numbers. Every single point on an infinitely long, continuous number line corresponds to exactly one Real number.
1.1 Prime Numbers & Factorization
Within the set of Natural numbers, we distinguish between Prime and Composite numbers. A Prime Number is a natural number strictly greater than 1 that has exactly two distinct positive divisors: 1 and itself. (e.g., 2, 3, 5, 7, 11, 13, 17...).
A Composite Number has more than two divisors. The Fundamental Theorem of Arithmetic states that every integer greater than 1 either is a prime itself or can be represented uniquely as a product of prime numbers. This is called Prime Factorization.
Example: Prime Factorization and LCM / GCD
Find the Greatest Common Divisor (GCD) and Lowest Common Multiple (LCM) of 60 and 72.
- Factorize 60:
60 = 6 × 10 = (2 × 3) × (2 × 5) = $2^2 \times 3^1 \times 5^1$ - Factorize 72:
72 = 8 × 9 = $(2 \times 2 \times 2) \times (3 \times 3)$ = $2^3 \times 3^2$ - Greatest Common Divisor (GCD): Take the lowest power of common prime factors.
GCD = $2^2 \times 3^1 = 4 \times 3 = \mathbf{12}$ - Lowest Common Multiple (LCM): Take the highest power of all prime factors present.
LCM = $2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = \mathbf{360}$
Note: Finding the LCM is the fundamental step for adding or subtracting fractions with different denominators.
1.2 Converting Repeating Decimals to Fractions
Proof: Repeating Decimals are Rational
Prove that $x = 2.141414...$ (written as $2.\overline{14}$) is a rational number.
- Let the variable $x$ represent the repeating decimal:
$x = 2.141414...$ - Because exactly 2 digits repeat ("14"), we multiply the entire equation by $10^2$ (which is 100) to shift the decimal point exactly one full repeating cycle to the right:
$100x = 214.141414...$ - Now, subtract the original equation from this new equation. The infinite repeating decimal tails will perfectly cancel each other out:
$100x = 214.141414...$
- $x = 2.141414...$
$99x = 212$ - Solve algebraically for $x$:
$x = \frac{212}{99}$
Because $2.1414...$ can be perfectly represented as the ratio of two integers (212 and 99), it is by definition a Rational number.
1.3 Scientific Notation (Standard Form)
Scientific notation is a method for expressing extremely large or extremely small numbers concisely. It is written in the form $A \times 10^n$, where $1 \le |A| < 10$ and $n$ is an integer.
- Large numbers: Move the decimal to the left, and $n$ is positive.
Example: $4,500,000 = 4.5 \times 10^6$. - Small numbers: Move the decimal to the right, and $n$ is negative.
Example: $0.000032 = 3.2 \times 10^{-5}$.
Challenge an IMAT Question!
Official Paper: 2023 - Q49
worked solution & explanation
Step 1 Calculate the discriminant ($\Delta = b^2 - 4ac$) to find intersections with the x-axis.
Step 2 $\Delta = (-6)^2 - 4(2)(5) = 36 - 40 = -4$.
Step 3 Since $\Delta \lt 0$ and the leading coefficient $a=2$ is positive, the parabola opens upwards and never touches the x-axis.
Challenge an IMAT Question!
Official Paper: 2018 - Q53
worked solution & explanation
Step 1 Multiply both sides by the denominator $ (x+1) $.
$ x + 4 = x(x+1) \implies x + 4 = x^2 + x $.
Step 2 Rearrange into standard quadratic form.
Subtract $ x $ from both sides: $ 4 = x^2 \implies x^2 - 4 = 0 $.
Step 3 Find roots and sum.
Roots are 2 and -2. Sum = $ 2 + (-2) = 0 $.
Part 2: Indices, Radicals, and Rationalization
Indices (exponents) and radicals (roots) are two sides of the exact same mathematical coin. An index denotes repeated multiplication, while a radical asks for the base that was multiplied. Mastery of the laws governing these operations is non-negotiable for algebra.
The Universal Laws of Indices
- Multiplication Law:
$a^m \cdot a^n = a^{m+n}$ - Division Law:
$a^m \div a^n = a^{m-n}$ - Power of a Power Law:
$(a^m)^n = a^{mn}$ - Power of a Product Law:
$(ab)^n = a^n b^n$ - Power of a Quotient Law:
$(\frac{a}{b})^n = \frac{a^n}{b^n}$ - Zero Index Rule:
$a^0 = 1$ (for $a \neq 0$) - Negative Index Rule:
$a^{-n} = \frac{1}{a^n}$ -
Fractional Index Rule (The Bridge to Radicals):
$$a^{\frac{m}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$$
2.1 Operations with Radicals
You can only add or subtract radicals if they have the exact same radicand (the number inside the root) and the same index. This is similar to combining like terms in algebra ($2x + 5x = 7x$).
Addition and Subtraction of Radicals
Correct: $3\sqrt{5} + 4\sqrt{5} = 7\sqrt{5}$
Incorrect: $\sqrt{2} + \sqrt{3} \neq \sqrt{5}$. (These are unlike terms and cannot be combined. $\sqrt{2} + \sqrt{3}$ is the simplest form).
However, always check if radicals can be simplified first to reveal hidden like terms. For example: $\sqrt{18} + \sqrt{50} = \sqrt{9 \times 2} + \sqrt{25 \times 2} = 3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$.
Challenge an IMAT Question!
Official Paper: 2022 - Q51
worked solution & explanation
Notice Align all exponents in the numerator to $10^8$ to simplify addition.
Step 1 Numerator: $(81 \times 10^8) - (1 \times 10^8) + (1200 \times 10^8) = 1280 \times 10^8$.
Step 2 Denominator: Convert to decimals for easier subtraction. $0.100 - 0.098 = 0.002 = 2 \times 10^{-3}$.
Step 3 Divide and apply the cube root: $\sqrt[3]{\frac{1280 \times 10^8}{2 \times 10^{-3}}} = \sqrt[3]{640 \times 10^{11}} = \sqrt[3]{64 \times 10^{12}}$.
Step 4 $\sqrt[3]{4^3 \times (10^4)^3} = 4 \times 10^4 = 40,000$.
Challenge an IMAT Question!
Official Paper: 2017 - Q55
worked solution & explanation
Step 1 Calculate the sum of the 3 items.
Sum = $ x/3 + x + (x + 6) = x/3 + 2x + 6 $. Create common denominator: $ (x + 6x)/3 + 6 = 7x/3 + 6 $.
Step 2 Divide sum by count (3) to find mean.
Mean = $ (7x/3 + 6) / 3 = 7x/9 + 2 $.
Step 3 Combine into a single fraction to match options.
$ 7x/9 + 18/9 = (7x+18)/9 $.
2.2 Rationalizing the Denominator
In mathematics, it is standard convention to remove irrational numbers (roots) from the denominator of a fraction. This process is called rationalization. We achieve this by multiplying the fraction by a strategic form of the number $1$.
-
Case 1: Single Term Denominator. If the denominator is a simple square root, multiply the numerator and the denominator by that exact square root.
Example: $$ \frac{7}{\sqrt{3}} = \frac{7}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{7\sqrt{3}}{(\sqrt{3})^2} = \frac{7\sqrt{3}}{3} $$ - Case 2: Binomial Denominator. If the denominator is an expression with two terms (like $a + \sqrt{b}$), you must multiply the numerator and denominator by its Conjugate (change the middle sign to $a - \sqrt{b}$). This exploits the difference of squares algebraic identity $(x-y)(x+y) = x^2 - y^2$ to eliminate the square roots entirely.
Example: Rationalize the Binomial Denominator
Simplify the expression: $\frac{2\sqrt{2}}{5 - \sqrt{7}}$
Step 1: Identify the conjugate of the denominator $(5 - \sqrt{7})$. The conjugate is $(5 + \sqrt{7})$.
Step 2: Multiply both the numerator and denominator by this conjugate.
$$ \frac{2\sqrt{2}}{5 - \sqrt{7}} \cdot \frac{5 + \sqrt{7}}{5 + \sqrt{7}} = \frac{2\sqrt{2}(5 + \sqrt{7})}{5^2 - (\sqrt{7})^2} = \frac{10\sqrt{2} + 2\sqrt{14}}{25 - 7} = \frac{10\sqrt{2} + 2\sqrt{14}}{18} $$
Step 3: Factor out common terms and simplify the fraction. We can divide the numerator and denominator by 2.
$$ \frac{2(5\sqrt{2} + \sqrt{14})}{18} = \frac{5\sqrt{2} + \sqrt{14}}{9} $$
Challenge an IMAT Question!
Official Paper: 2021 - Q54
worked solution & explanation
Step 1 Multiply the entire inequality by the Least Common Multiple (LCM) of 2 and 3, which is 6, to clear denominators: $3(2x+3) - 4(x+1) \lt 12x$.
Step 2 Expand brackets and collect like terms: $6x + 9 - 4x - 4 \lt 12x \implies 2x + 5 \lt 12x$.
Step 3 Isolate x: $5 \lt 10x \implies x \gt 1/2$.
Challenge an IMAT Question!
Official Paper: 2017 - Q60
worked solution & explanation
Concept Permutation with repetition formula: $ \frac{N!}{n_1! n_2! \dots} $.
Step 1 Count total and repeating letters: Total $ N = 5 $. Two 'A's ($ n_1 = 2 $), Two 'B's ($ n_2 = 2 $), One 'C'.
Step 2 Calculate: $ \frac{5!}{2! \cdot 2! \cdot 1!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30 $.
2.3 Advanced Technique: Unnesting Double Radicals
An expression nested inside another root, like $\sqrt{A \pm \sqrt{B}}$, can sometimes be simplified into the much cleaner form $\sqrt{a} \pm \sqrt{b}$. The trick relies directly on reversing the binomial expansion formula: $(\sqrt{a} \pm \sqrt{b})^2 = a + b \pm 2\sqrt{ab}$.
The Golden Rule for Double Radicals
To simplify an expression into the form $\sqrt{X \pm 2\sqrt{Y}}$, you absolutely must have a coefficient of exactly 2 in front of the inner root. Once you have that "2", you simply find two numbers $a$ and $b$ such that:
- The sum $a + b = X$
- The product $a \cdot b = Y$
Once found, the simplified form is $\sqrt{a} \pm \sqrt{b}$. Crucial Note: Always write the larger number first if dealing with subtraction to ensure a positive result!
Example: Unnesting $\sqrt{11 - \sqrt{112}}$
- Create the "2": We must extract a 4 from the inner root 112 because $\sqrt{4}=2$.
$\sqrt{112} = \sqrt{4 \times 28} = 2\sqrt{28}$.
The expression is now rewritten as: $\sqrt{11 - 2\sqrt{28}}$. - Find the numbers: We need two numbers that add up to 11 and multiply together to make 28.
Let's list factor pairs of 28: (1, 28), (2, 14), (4, 7).
The pair $4$ and $7$ works perfectly because $4 + 7 = 11$. - Apply the rule: Write the roots, placing the larger number first.
$\sqrt{7} - \sqrt{4}$ - Final Simplification: We know that $\sqrt{4} = 2$.
The final answer is $\mathbf{\sqrt{7} - 2}$.
Part 3: Comprehensive Polynomial Expansion
Expansion is the process of multiplying out brackets. Memorizing these standard expansions saves critical time during the exam. They form the foundational vocabulary of all advanced algebra.
The Ultimate Algebraic Expansion Cheat Sheet
- Quadratic Expansions:
Perfect Square (Sum): $(a + b)^2 = a^2 + 2ab + b^2$
Perfect Square (Difference): $(a - b)^2 = a^2 - 2ab + b^2$
Difference of Squares: $(a + b)(a - b) = a^2 - b^2$
Square of a Trinomial: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
- Cubic Expansions:
Perfect Cube (Sum): $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
Perfect Cube (Difference): $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
Sum of Cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
Difference of Cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
3.1 Geometric Visualization of Expansion
Algebraic identities are not arbitrary rules; they describe fundamental geometric realities. Let us visualize why $(a+b)^2 = a^2 + 2ab + b^2$ by looking at the area of a square.
Geometric Proof of the Perfect Square Expansion
3.2 Pascal's Triangle for Higher Powers
If you need to expand $(a+b)^4$ or higher, memorizing formulas becomes impractical. Pascal's Triangle provides the exact coefficients needed for any binomial expansion $(a+b)^n$.
To build the triangle, start with 1 at the top. Every number below is the sum of the two numbers directly above it. The row index $n$ (starting at $n=0$ for the top row) corresponds to the power of the binomial.
Pascal's Triangle
Challenge an IMAT Question!
Official Paper: 2025 - Q51
worked solution & explanation
Concept Recognize the perfect square trinomial pattern: $x^2 - 2xy + y^2 = (x-y)^2$.
Step 1 Identify the square roots of the first and last terms: $\sqrt{a^2} = a$ and $\sqrt{4b^2} = 2b$.
Step 2 Check the middle term: $-2(a)(2b) = -4ab$. It matches perfectly.
Challenge an IMAT Question!
Official Paper: 2023 - Q50
worked solution & explanation
Step 1 Let initial sugar be $x$ and flour be $5x$. Equation for first addition: $\frac{x + 1}{5x + 2} = \frac{2}{5}$.
Step 2 Cross-multiply: $5(x + 1) = 2(5x + 2) \implies 5x + 5 = 10x + 4 \implies x = 0.2$ kg.
Step 3 Current amounts: Sugar = $0.2 + 1 = 1.2$ kg. Flour = $5(0.2) + 2 = 3.0$ kg.
Step 4 Add second batch: New Sugar = $1.2 + 1 = 2.2$. New Flour = $3.0 + 2 = 5.0$.
Step 5 Find the final ratio: $2.2 / 5.0 = 22 / 50 = 11 / 25$.
| Stage | Sugar (kg) | Flour (kg) | Ratio (S:F) |
|---|---|---|---|
| 1. Initial (x = 0.2) | 0.2 | 1.0 | 1 : 5 |
| 2. First Addition (+1S, +2F) | 1.2 | 3.0 | 2 : 5 (1.2:3) |
| 3. Second Addition (+1S, +2F) | 2.2 | 5.0 | 11 : 25 (2.2:5) |
Challenge an IMAT Question!
Official Paper: 2018 - Q60
worked solution & explanation
Step 1 Distribute the exponent into the bracket.
$ (3 \times 10^3)^3 = 3^3 \times 10^{3 \times 3} = 27 \times 10^9 $.
Step 2 Multiply coefficients and add base-10 exponents separately.
$ (27 \times 2) \times 10^{9 + (-5)} = 54 \times 10^4 $.
Step 3 Format into proper scientific notation ($ A \times 10^B $ where $ 1 \le A \lt 10 $).
$ 54 \times 10^4 = 5.4 \times 10^5 $.
Part 4: The Art and Strategy of Factoring
Factoring takes an expanded polynomial and breaks it down into a product of simpler parentheses. It is the exact reverse of expansion and is the most critical skill for solving equations. When faced with an expression to factor, you must follow this strict hierarchy of attack:
-
1. Greatest Common Factor (GCF): Always check for this first before applying any complex formulas.
Solve: $12x^3y^2 - 18x^2y^3$
The GCF of the numbers is 6. For variables, take the lowest power: $x^2$ and $y^2$.
Result: $6x^2y^2(2x - 3y)$ -
2. Recognize Standard Formulas: Look for patterns from Part 3.
Difference of Squares: $49x^2 - 25 = (7x - 5)(7x + 5)$
Sum of Cubes: $8x^3 + 27 = (2x)^3 + 3^3 = (2x + 3)(4x^2 - 6x + 9)$ -
3. Factoring by Grouping (4 or more terms): Group terms strategically to extract a common binomial bracket.
Expression: $x^3 + 2x^2 - 4x - 8$
Group first two and last two: $(x^3 + 2x^2) - (4x + 8)$
Factor each group: $x^2(x + 2) - 4(x + 2)$
Extract common bracket: $(x + 2)(x^2 - 4)$
Factor the remaining difference of squares: $\mathbf{(x + 2)(x - 2)(x + 2) = (x + 2)^2(x - 2)}$ -
4. The AC Method / Cross Method for Quadratics:
When factoring a trinomial $ax^2 + bx + c$ where the leading coefficient $a \neq 1$, trial and error is frustrating. The Cross Method (AC Method) is a visual algorithm that guarantees success. You pair factors of $a$ and factors of $c$, cross-multiply them, and ensure their sum matches the middle term $b$.
Factor: $6x^2 + 11x - 10$
Since the sum matches the middle term $11x$, we read the factors horizontally row by row.
Result: $(2x + 5)(3x - 2)$
4.1 The Factor Theorem
If you are dealing with a polynomial $P(x)$ of degree 3 or higher, and grouping doesn't work, you must use the Factor Theorem. The theorem states: If you plug a number $c$ into the polynomial and the result is zero (i.e., $P(c) = 0$), then $(x - c)$ is guaranteed to be a factor.
Example: Factor $P(x) = x^3 - 7x + 6$
Step 1: Guess values for $x$ that are divisors of the constant term (6). Let's test $\pm 1, \pm 2, \pm 3$.
Test x=1: $P(1) = 1^3 - 7(1) + 6 = 1 - 7 + 6 = 0$.
Since $P(1) = 0$, the Factor Theorem tells us that $(x - 1)$ is a factor.
Step 2: To find the remaining quadratic factor, divide the original polynomial by $(x - 1)$ using Polynomial Long Division (which we will cover in the next section).
The division yields $x^2 + x - 6$.
Step 3: Factor the resulting quadratic: $x^2 + x - 6 = (x + 3)(x - 2)$.
Final complete factorization: $P(x) = (x - 1)(x - 2)(x + 3)$
Challenge an IMAT Question!
Official Paper: 2020 - Q56
worked solution & explanation
Step 1 Convert all terms to a common base of 2.
$ 8^{2x+3} = (2^3)^{2x+3} = 2^{6x+9} $.
$ 1/4^{3x} = 4^{-3x} = (2^2)^{-3x} = 2^{-6x} $.
Step 2 Combine the left side by adding exponents: $2^{6x+9 - 6x} = 2^9$.
Step 3 Equate the exponents of both sides: $2^9 = 2^{x+3} \implies 9 = x + 3 \implies x = 6$.
Challenge an IMAT Question!
Official Paper: 2015 - Q58
worked solution & explanation
Concept Avoid manual large multiplication. Use Difference of Squares: $a^2 - b^2 = (a+b)(a-b)$.
Step 1 First part: $27^2 - 23^2 = (27+23)(27-23) = (50)(4) = 200$.
Step 2 Second part: $14^2 - 6^2 = (14+6)(14-6) = (20)(8) = 160$.
Step 3 Add them: $200 + 160 = 360$.
Challenge an IMAT Question!
Official Paper: 2021 - Q53
worked solution & explanation
Step 1 Factorize the quadratic denominator. Two numbers that multiply to -2 and add to 1 are +2 and -1. Denominator becomes $(x+2)(x-1)$.
Step 2 Rewrite fraction and cancel out common terms: $\frac{(x+2)(x+2)}{(x+2)(x-1)}$.
Step 3 Simplify to $\frac{x+2}{x-1}$.
Part 5: Solving Equations & Polynomial Division
An equation $f(x) = 0$ is a question: "At what exact points does the graph of this mathematical function cross the horizontal x-axis?" Solving an equation is finding those roots.
5.1 Quadratic Equations & The Discriminant
When a quadratic equation $ax^2 + bx + c = 0$ cannot be factored easily, the Quadratic Formula is the universal tool.
The value inside the square root, $D = b^2 - 4ac$, is called the Discriminant. It fundamentally controls the nature of the roots by determining what happens when you add or subtract the square root.
Visualizing the Discriminant's Effect on the Parabola
Case 1: $D > 0$
$\sqrt{Positive}$ yields a real value. You step left and right from the vertex.
Two Real Roots.
Case 2: $D = 0$
$\pm \sqrt{0}$ adds nothing. The vertex sits exactly on the axis.
One Repeated Root.
Case 3: $D < 0$
$\sqrt{Negative}$ is not a Real number. The graph never touches the axis.
No Real Roots.
Challenge an IMAT Question!
Official Paper: 2025 - Q53
worked solution & explanation
Step 1 Rearrange the linear equation to solve for $x$: $3x = 3 - a \implies x = (3-a)/3$.
Step 2 Since there is no division by zero or any variable in the denominator, this expression yields a valid real number for $x$ regardless of the value of $a$.
Challenge an IMAT Question!
Official Paper: 2025 - Q54
worked solution & explanation
Step 1 Isolate $x$ by dividing: $x = 5 / (a+3)$.
Step 2 A mathematical expression is impossible (undefined) if its denominator is zero.
Step 3 Set denominator to zero: $a+3 = 0 \implies a = -3$.
Challenge an IMAT Question!
Official Paper: 2024 - Q51
worked solution & explanation
Concept Analyze the denominator's sign to simplify rational inequalities rapidly.
Step 1 Analyze the denominator: $x^2 + |4x+3|$. For any real number, $x^2 \ge 0$ and absolute values are $\ge 0$. Thus, the entire denominator is strictly positive ($\gt 0$).
Step 2 Since the denominator is positive, the entire fraction is $\ge 0$ only if the numerator is $\ge 0$.
Step 3 Solve numerator: $4 - 3x \ge 0 \implies 4 \ge 3x \implies x \le 4/3$.
Challenge an IMAT Question!
Official Paper: 2020 - Q57
worked solution & explanation
Step 1 Find total values. Total students = $39 + 36 = 75$. Overall mean = $3.2$. Total meals = $75 \times 3.2 = 240$.
Step 2 Find total male meals: $39 \times 2 = 78$ meals.
Step 3 Subtract to find female meals: $240 - 78 = 162$ meals.
Step 4 Calculate female mean: $162 / 36 = 4.5 = 4\frac{1}{2}$.
| Group | Count (N) | Mean | Total Meals (Count × Mean) |
|---|---|---|---|
| Males | 39 | 2.0 | 78 |
| Females | 36 | x | 36 × x = 162 |
| Overall | 75 | 3.2 | 240 |
Challenge an IMAT Question!
Official Paper: 2015 - Q55
worked solution & explanation
Step 1 Solve second equation for y (Substitution method): $2x - y = 5 \implies y = 2x - 5$.
Step 2 Substitute into first equation: $x + 3(2x - 5) = 13 \implies x + 6x - 15 = 13 \implies 7x = 28 \implies x = 4$.
Step 3 Find y and sum: $y = 2(4) - 5 = 3$. Thus $x+y = 4+3 = 7$.
5.2 Biquadratic Equations
Equations of the form $ax^4 + bx^2 + c = 0$ appear terrifying because of the 4th power, but they are secretly quadratics in disguise. You solve them using a mathematical technique called substitution.
Example: Solve $x^4 - 13x^2 + 36 = 0$
- Let a new variable, $y = x^2$. Consequently, $y^2 = x^4$.
- Substitute $y$ into the original equation to transform it:
$y^2 - 13y + 36 = 0$ - This is now a standard quadratic! Factor it:
$(y - 9)(y - 4) = 0 \implies y = 9$ or $y = 4$ - Crucial Step: Reverse the substitution. You are solving for $x$, not $y$.
If $y = 9 \implies x^2 = 9 \implies \mathbf{x = \pm 3}$
If $y = 4 \implies x^2 = 4 \implies \mathbf{x = \pm 2}$ - The equation has four distinct real solutions: $x = -3, -2, 2, 3$.
5.3 Polynomial Long Division
When applying the Factor Theorem, you must divide a larger polynomial by a known linear factor to find the remaining polynomial. Polynomial Long Division works exactly like elementary numerical long division.
Example: Divide $(2x^3 - 5x^2 + 8x - 4)$ by $(x - 1)$
Therefore, the result is: $\mathbf{2x^2 - 3x + 5 + \frac{1}{x-1}}$
Challenge an IMAT Question!
Official Paper: 2023 - Q55
worked solution & explanation
Notice Rather than converting to base 2 immediately, factoring out the smallest term is much faster.
Step 1 Factor numerator: $8^x(2^x + 1)$. Factor denominator: $2^x(2^x + 1)$.
Step 2 Cancel the common binomial factor $ (2^x+1) $: expression becomes $\sqrt{\frac{8^x}{2^x}}$.
Step 3 Simplify using exponent rules: $\sqrt{4^x} = \sqrt{(2^2)^x} = \sqrt{2^{2x}} = 2^x$.
Challenge an IMAT Question!
Official Paper: 2016 - Q55
worked solution & explanation
Step 1 Factor the denominator of the first term to find the common denominator: $ x^2 - 1 = (x-1)(x+1) $.
Step 2 Multiply the second fraction by $ (x+1) $ to match denominators, then subtract.
$ \frac{2}{(x-1)(x+1)} - \frac{1(x+1)}{(x-1)(x+1)} = \frac{2 - x - 1}{(x-1)(x+1)} $.
Step 3 Simplify numerator: $ 1 - x $. Cancel with denominator term $ (x-1) $ by factoring out a $-1$: $ \frac{-(x-1)}{(x-1)(x+1)} = \frac{-1}{x+1} $.
5.4 Rational and Radical Equations (The Threat of Extraneous Roots)
WARNING: Extraneous Solutions (Ghost Roots)
Certain algebraic operations—specifically multiplying by a variable expression or squaring both sides of an equation—can mathematically create "ghosts". These are numbers that perfectly solve your final algebraic step, but cause a catastrophic error (like division by zero) or a logical impossibility in the original equation.
You must check your answers in the original equation!
Rational Example
Solve: $\frac{x}{x-2} - \frac{2}{x+1} = \frac{6}{x^2 - x - 2}$
- 1. Factor Denominators: The right side $x^2 - x - 2$ factors to $(x-2)(x+1)$. Thus, the LCD is $(x-2)(x+1)$.
Crucial restrictions: $x \neq 2$ and $x \neq -1$. - 2. Multiply by LCD to clear fractions:
$x(x+1) - 2(x-2) = 6$ - 3. Solve the resulting equation:
$x^2 + x - 2x + 4 = 6$
$x^2 - x - 2 = 0$
$(x-2)(x+1) = 0 \implies x=2, x=-1$ - 4. Check against restrictions: Both mathematical answers violate the initial constraints (they cause division by zero).
- Final Result: No Real Solution.
Radical Example
Solve: $\sqrt{x} = -3$
- 1. Conceptual Check: The principal square root function $\sqrt{x}$ always yields a positive number or zero. It can never equal -3. The equation is logically false. Let's see what happens if we blindly calculate.
- 2. Square both sides:
$(\sqrt{x})^2 = (-3)^2$ - 3. Solve:
$x = 9$ - 4. Check the ghost root: Plug $x=9$ back into the original equation:
$\sqrt{9} = 3$. But the equation demands it equal $-3$.
$3 \neq -3$. - Final Result: No Real Solution.
Challenge an IMAT Question!
Official Paper: 2020 - Q53
worked solution & explanation
Concept Rationalize the denominator by multiplying top and bottom by the conjugate.
Step 1 The conjugate of $2 + \sqrt{3}$ is $2 - \sqrt{3}$.
Step 2 Multiply the denominator using difference of squares: $(2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.
Step 3 Multiply the numerator: $2(2-\sqrt{3}) = 4 - 2\sqrt{3}$. Since the denominator is 1, this is the final answer.
Challenge an IMAT Question!
Official Paper: 2014 - Q56
worked solution & explanation
Step 1 Adjust exponent to even number before rooting: $\sqrt{1.6 \times 10^7} = \sqrt{16 \times 10^6} = 4 \times 10^3$.
Step 2 Expand the squared term: $(1.2 \times 10^3)^2 = 1.44 \times 10^6$.
Step 3 Combine coefficients: $(8 / 4) \times 1.44 = 2 \times 1.44 = 2.88$.
Step 4 Combine base-10 exponents: $10^{-5} / 10^3 \times 10^6 = 10^{-5 - 3 + 6} = 10^{-2}$.
Part 6: Flawless Inequalities & The Wavy Curve Method
The Golden Rule of Inequalities
If you multiply or divide both sides of an inequality by a negative number, you MUST flip the inequality sign.
Multiplying by a negative reflects numbers across zero on the number line. While $5 > 2$ is true, multiplying by -1 reflects them to $-5$ and $-2$. Since $-5$ is further left, $-5 < -2$. The sign must flip.
6.1 Higher-Degree Polynomial Inequalities (The Wavy Curve Method)
When dealing with inequalities involving polynomials of degree 3 or higher, or complex rational fractions, calculating test values for every single interval takes too much time. The Wavy Curve Method (Sign Chart) is a powerful visual algorithm to solve these instantly.
Example: Solve $(x-1)(x+2)(x-4) > 0$
- Identify the Roots: Find where each factor equals zero. The roots are $x=1, x=-2,$ and $x=4$.
- Plot on a Number Line: Place these roots in ascending order on a number line.
- Draw the Wave: Start from the top right (positive infinity). Since all $x$ terms are positive ($+x$, not $-x$), the function starts positive. Draw a continuous wave that passes through each root, alternating from positive (above the line) to negative (below the line).
Select the correct regions: The inequality asks for values $> 0$ (strictly positive). We look at our wave and select the regions where the curve is above the number line.
Final Answer: $-2 < x < 1 \text{ or } x > 4$
Challenge an IMAT Question!
Official Paper: 2025 - Q48
worked solution & explanation
Concept Domain checking is required for even roots.
Step 1 Determine the domain of the root: $2x \ge 0 \implies x \ge 0$.
Step 2 Since $x \ge 0$, the right side $1+x$ is guaranteed to be positive. Therefore, square both sides.
Step 3 $2x \lt 1 + 2x + x^2 \implies x^2 > -1$. Since $x^2$ is always $\ge 0$, this is universally true for the defined domain.
Challenge an IMAT Question!
Official Paper: 2024 - Q48
worked solution & explanation
Concept Apply the power of a power rule: $(x^a)^b = x^{a \cdot b}$.
| Rule Name | Formula | Application Example |
|---|---|---|
| Product of Powers | $x^a \cdot x^b = x^{a+b}$ | $2^3 \cdot 2^2 = 2^5 = 32$ |
| Power of a Power | $(x^a)^b = x^{ab}$ | $(512^{1/3})^{1/2} = 512^{1/6}$ |
| Fractional Exponent | $x^{m/n} = \sqrt[n]{x^m}$ | $8^{2/3} = \sqrt[3]{64} = 4$ |
Step 1 Multiply exponents: $(512^{1/3})^{1/2} = 512^{(1/3) \times (1/2)} = 512^{1/6}$.
Step 2 Recognize $512$ as a power of 2 ($512 = 2^9$): $(2^9)^{1/6} = 2^{9/6} = 2^{3/2}$.
Step 3 Convert back to radical: $2^{3/2} = \sqrt{2^3} = \sqrt{8} = 2\sqrt{2}$.
Challenge an IMAT Question!
Official Paper: 2015 - Q53
worked solution & explanation
Step 1 Convert all terms on the left side into log format.
$ 2\log_{10}(x) = \log_{10}(x^2) $. Constant $3 = \log_{10}(10^3) = \log_{10}(1000)$.
Step 2 Use log subtraction rule to combine: $\log_{10}(x^2) - \log_{10}(1000) = \log_{10}(x^2 / 1000)$.
Step 3 Equate to right side and remove logs: $\log_{10}(x^2 / 1000) = \log_{10}(y) \implies y = x^2 / 1000$.
6.2 Fractional Inequalities: The Squaring Method
Consider the inequality $\frac{x-1}{x+3} \ge 0$. You cannot multiply both sides by the denominator $(x+3)$ because you do not know if $(x+3)$ is positive or negative. Depending on the unknown value of $x$, you might need to flip the inequality sign, which makes multiplication invalid.
Method: Multiplying by the Square of the Denominator
While we don't know the sign of $(x+3)$, we are absolutely certain that the square, $(x+3)^2$, is strictly positive (for any real number $x \neq -3$). Therefore, we can safely multiply both sides of the inequality by $(x+3)^2$ without ever violating the Golden Rule!
Solve: $\frac{x-1}{x+3} \ge 0$
1. State restriction: $x \neq -3$
2. Multiply by $(x+3)^2$:
$\frac{x-1}{x+3} \cdot \mathbf{(x+3)^2} \ge 0 \cdot \mathbf{(x+3)^2}$
3. Simplify (one $(x+3)$ cancels out):
$(x-1)(x+3) \ge 0$
4. Solve this standard quadratic (roots are 1 and -3, opens up, we want $\ge 0$):
$x \le -3 \text{ or } x \ge 1$
5. Apply the restriction from step 1 ($x \neq -3$):
$\mathbf{x < -3 \text{ or } x \ge 1}$
6.3 Absolute Value Equations and Inequalities
The absolute value $|X|$ geometrically represents the distance of a number $X$ from zero on the number line. Because distance is always positive, absolute value equations and inequalities naturally split into two separate scenarios.
"Sandwich" Distance
Distance is less than $a$. Must stay close to zero.
$$ |X| < a \implies -a < X < a $$
"Split" Distance
Distance is greater than $a$. Must be far from zero.
$$ |X| > a \implies X < -a \text{ or } X > a $$
Challenge an IMAT Question!
Official Paper: 2024 - Q49
worked solution & explanation
Step 1 Substitute $x = 2$ into the function: $f(2) = \log_2(2^2 + 12) = \log_2(4 + 12) = \log_2(16)$.
Step 2 Evaluate the logarithm. Since $ 2^4 = 16 $, we know $\log_2(16) = 4$.
Notice The question explicitly asks for the reciprocal of $f(2)$, not the value itself. The reciprocal of 4 is 1/4.
Challenge an IMAT Question!
Official Paper: 2019 - Q57
worked solution & explanation
Step 1 Prime factorize each number.
$ 360 = 2^3 \cdot 3^2 \cdot 5^1 $.
$ 500 = 2^2 \cdot 5^3 $.
$ 700 = 2^2 \cdot 5^2 \cdot 7^1 $.
Step 2 For the Highest Common Factor (HCF), take the smallest power of all common prime factors. The common primes are 2 and 5.
Smallest power of 2 is $2^2$. Smallest power of 5 is $5^1$.
Part 7: Logarithms & Exponential Equations
If an exponent answers the question "What is the base multiplied by itself $y$ times?", a logarithm answers the reverse: "To what power $y$ must we raise the base $b$ to get the number $x$?"
The Fundamental Definition
$$ y = \log_b(x) \iff b^y = x $$
(Where base $b > 0$, $b \neq 1$, and argument $x > 0$)
The Laws of Logarithms
Because logarithms are the exact inverse of exponents, their laws are direct translations of the Laws of Indices.
- Product Law: $\log_b(xy) = \log_b(x) + \log_b(y)$
(Multiplication inside becomes addition outside). - Quotient Law: $\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)$
(Division inside becomes subtraction outside). - Power Law: $\log_b(x^n) = n \cdot \log_b(x)$
(Powers drop down to the front as multipliers). - Change of Base Formula: $\log_b(x) = \frac{\log_c(x)}{\log_c(b)}$
(Allows evaluation on calculators using natural log (ln) or base 10).
Example: Solving an Exponential Equation using Logarithms
Solve for $x$: $5^{2x-1} = 7$
- Since the bases (5 and 7) cannot be made equal, we must "drop" the exponent down. Take the natural logarithm ($\ln$) of both sides:
$\ln(5^{2x-1}) = \ln(7)$ - Use the Power Law of Logarithms to bring the exponent to the front:
$(2x - 1) \cdot \ln(5) = \ln(7)$ - Divide both sides by $\ln(5)$ (which is just a constant number):
$2x - 1 = \frac{\ln(7)}{\ln(5)}$ - Isolate $x$ using basic algebra:
$2x = \frac{\ln(7)}{\ln(5)} + 1$
$\mathbf{x = \frac{\frac{\ln(7)}{\ln(5)} + 1}{2}}$
Challenge an IMAT Question!
Official Paper: 2023 - Q53
worked solution & explanation
Step 1 Solve the first inequality: $100 - 98 \le 99x - 97x \implies 2 \le 2x \implies x \ge 1$.
Step 2 Solve the second inequality: $95x - 93x \gt 94 - 96 \implies 2x \gt -2 \implies x \gt -1$.
Step 3 Find the intersection (AND condition). A number that is $\ge 1$ AND $\gt -1$ is simply $x \ge 1$.
Challenge an IMAT Question!
Official Paper: 2019 - Q58
worked solution & explanation
Step 1 Clear denominators by multiplying all terms by $ x(x-2) $.
$ 3(x-2) + 2x = x(x-2) \implies 3x - 6 + 2x = x^2 - 2x $.
Step 2 Rearrange into standard quadratic form ($ ax^2 + bx + c = 0 $).
$ 5x - 6 = x^2 - 2x \implies x^2 - 7x + 6 = 0 $.
Concept Vieta's Formula: The sum of roots of a quadratic equation is $ -b/a $.
Step 3 Sum = $ -(-7)/1 = 7 $.
Challenge an IMAT Question!
Official Paper: 2019 - Q59
worked solution & explanation
Step 1 Convert all bases to 2.
Numerator: $ (2^3)^{2n} \times (2^2)^n = 2^{6n} \times 2^{2n} = 2^{8n} $.
Step 2 Divide by denominator $ 2^n $ by subtracting exponents.
$ 2^{8n} / 2^n = 2^{8n - n} = 2^{7n} $.
Challenge an IMAT Question!
Official Paper: 2017 - Q57
worked solution & explanation
Concept Logarithm rules: $ \log(A/B) = \log A - \log B $ and $ \log(AB) = \log A + \log B $.
Step 1 Apply division rule: $ \log_{10}(14/3) = \log_{10}(14) - \log_{10}(3) $.
Step 2 Factor 14 and apply multiplication rule: $ \log_{10}(2 \times 7) = \log_{10}(2) + \log_{10}(7) $.
Step 3 Substitute variables: $ y + x - z = x+y-z $.
Challenge an IMAT Question!
Official Paper: 2014 - Q58
worked solution & explanation
Step 1 A 20% reduction means the sale price is 80% (or 0.80) of the original price $P$.
Step 2 Set up equation: $0.80 \times P = 32$.
Step 3 Solve: $P = 32 / 0.8 = 320 / 8 = 40$.
Challenge an IMAT Question!
Official Paper: 2013 - Q58
worked solution & explanation
Step 1 Addition of logs equals multiplication of arguments: $ \ln \left( (\frac{x^2}{4y}) \cdot (xy) \cdot (8) \right) $
Step 2 Simplify fraction inside log: $\ln \left( \frac{8x^3y}{4y} \right) = \ln(2x^3)$.
Step 3 Split logs and move exponent: $\ln(2) + \ln(x^3) = \ln(2) + 3\ln(x)$.
Challenge an IMAT Question!
Official Paper: 2013 - Q59
worked solution & explanation
Step 1 Solve quadratic inequality: $12 - x^2 \gt 8 \implies 4 \gt x^2 \implies -2 \lt x \lt 2$.
Step 2 Solve linear inequality: $2x + 3 \ge 5 \implies 2x \ge 2 \implies x \ge 1$.
Step 3 Find intersection (AND condition): overlap of $-2 \lt x \lt 2$ and $x \ge 1$ is $1 \le x \lt 2$.
Challenge an IMAT Question!
Official Paper: 2012 - Q76
worked solution & explanation
Step 1 Move coefficients to exponents: $\ln(x^2y) - \ln((xy)^2) + \ln(y^3)$.
Step 2 Combine using log properties: $\ln \left( \frac{x^2y \cdot y^3}{x^2y^2} \right)$.
Step 3 Simplify fraction: $x^2$ cancels. $y^4 / y^2 = y^2$. Result is $\ln(y^2) = 2\ln(y)$.
Part 8: Mathematical Translation (Word Problems)
A significant portion of the IMAT tests your ability to translate real-world English sentences into abstract algebraic equations. This requires mapping verbal relationships directly to mathematical operators.
8.1 Translation Dictionary
| English Phrase | Algebraic Translation |
|---|---|
| "Is", "Yields", "Results in" | = |
| "Sum of", "More than", "Increased by" | + |
| "Difference", "Less than", "Decreased by" | - |
| "Product", "Of" (e.g., half of x) | × or · |
| "Quotient", "Per", "Out of" | ÷ or / |
| "x is 5 less than y" | x = y - 5 (Watch the order!) |
8.2 Classic IMAT Problem Types
Age Problems
"A father is currently 3 times as old as his son. In 12 years, he will be exactly twice as old as his son. Find their current ages."
- Setup Variables: Let son's current age = $x$. Thus, father's current age = $3x$.
- Fast Forward 12 Years:
Son's age = $x + 12$
Father's age = $3x + 12$ - Translate the relationship:
$\text{Father's future age} = 2 \times (\text{Son's future age})$
$(3x + 12) = 2(x + 12)$ - Solve:
$3x + 12 = 2x + 24$
$x = 12$ - Result: Son is 12, Father is 36.
Distance-Speed-Time
"A train travels 120km at speed V. If the speed were increased by 20km/h, the journey would take 1 hour less. Find V."
- Recall Formula: $Time = \frac{Distance}{Speed}$
- Original Time: $T_1 = \frac{120}{V}$
- New Time: $T_2 = \frac{120}{V + 20}$
- Translate "1 hour less":
$T_1 - T_2 = 1$
$\frac{120}{V} - \frac{120}{V + 20} = 1$ - Solve (Multiply by LCD):
$120(V+20) - 120V = V(V+20)$
$2400 = V^2 + 20V$
$V^2 + 20V - 2400 = 0 \implies (V-40)(V+60) = 0$ - Result: Speed is 40 km/h (reject -60).
Part 9: Ultimate IMAT Mathematical Mastery (50 Questions)
This comprehensive diagnostic test covers every single concept taught in this module. From identifying prime numbers to expanding complex binomials, unwrapping double radicals, executing polynomial long division, and dissecting tricky word problems.
Instructions: Answer all 50 questions. Take your time. Once submitted, detailed step-by-step explanations will be revealed for every single question to help you analyze your mistakes.
Your Diagnostic Score
Review the rigorous explanations below.
True mastery comes from understanding exactly why an answer is correct.