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IMAT Mathematics · Formula sheet

Equations and Inequalities formulas

14 formulas and 16 common traps for IMAT Mathematics Equations and Inequalities, grouped by subtopic.

Full notes with worked examples

Expanding, Factorising and Algebraic Fractions

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Expanding brackets and the special products

Special products

(a±b)2=a2±2ab+b2(a+b)(a−b)=a2−b2(a \pm b)^2 = a^2 \pm 2ab + b^2 \qquad (a+b)(a-b) = a^2 - b^2

Factorising: common factors, grouping and trinomials

Monic trinomial

x2+sx+p=(x+m)(x+n)wherem+n=s,  mn=px^2 + sx + p = (x + m)(x + n) \quad\text{where}\quad m + n = s,\; mn = p

Simplifying and combining algebraic fractions

Adding fractions

ab±cd=ad±bcbd\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}

Polynomial division and the factor theorem

Remainder theorem

p(x)=(x−a) q(x)+p(a)p(x) = (x - a)\,q(x) + p(a)
  • q(x)q(x)the quotient, one degree lower than p(x)p(x)
  • p(a)p(a)the remainder, a number

Common traps

The square of a sum is not the sum of the squares

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, not a2+b2a^2 + b^2. Dropping the middle term 2ab2ab is the commonest expansion error, and IMAT offers the result as an option.

A sum of two squares does not factorise

x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3), but x2+9x^2 + 9 has no real factors: (x+3)(x−3)(x + 3)(x - 3) expands to x2−9x^2 - 9, and (x+3)2(x + 3)^2 has a middle term.

Cancel factors, never terms

x+5x+7\frac{x + 5}{x + 7} is not 57\frac{5}{7}: the xx is a term added on, not a factor of the whole top and bottom. Factorise first, then cancel only whole brackets that multiply everything.

For the factor x + a, test x = minus a

x+ax + a is zero when x=−ax = -a, so that is the value to substitute. Substituting x=ax = a answers a different question, and its result is always among the options.

Linear Equations, Parameters and Systems

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Linear equations and equations with a parameter

Solution of ax = b

ax=b,  a≠0  ⇒  x=baax = b,\; a \ne 0 \;\Rightarrow\; x = \frac{b}{a}

Systems of two linear equations

Two equations in two unknowns

{a1x+b1y=c1a2x+b2y=c2\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}

Common traps

Zero coefficient: check the right-hand side

When the coefficient of xx is zero, the equation is not automatically impossible. If the right-hand side is also zero, every xx works. Only a zero coefficient with a non-zero right-hand side gives no solution.

Proportional coefficients do not always mean no solution

If the xx and yy coefficients are in proportion, the lines are parallel. They have no common point only if the constants break the proportion; if the constants follow it too, the lines coincide and there are infinitely many solutions.

Quadratic Equations, the Discriminant and the Roots

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Solving quadratic equations and the discriminant

Quadratic formula

x=−b±Δ2a,Δ=b2−4acx = \frac{-b \pm \sqrt{\Delta}}{2a}, \qquad \Delta = b^2 - 4ac
  • a,b,ca, b, cthe coefficients of x2x^2, xx and the constant
  • Δ\Deltathe discriminant

Sum and product of the roots

x1+x2=−bax1x2=cax_1 + x_2 = -\frac{b}{a} \qquad x_1 x_2 = \frac{c}{a}

Common traps

Never divide both sides by x

Dividing 3x2=12x3x^2 = 12x by xx gives only x=4x = 4 and loses x=0x = 0. Bring everything to one side and factorise out xx instead.

The sum of the roots is minus b over a

Expanding a(x−x1)(x−x2)a(x - x_1)(x - x_2) gives −a(x1+x2)-a(x_1 + x_2) as the xx coefficient, so the sum is −ba-\frac{b}{a}, not ba\frac{b}{a}. And check Δ≥0\Delta \ge 0 first: x2+2x+5=0x^2 + 2x + 5 = 0 has no real roots, so it has no real roots to add.

Linear, Quadratic, Fractional and Absolute Value Inequalities

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Linear inequalities and systems of inequalities

Multiplying by a negative number

a<b  and  c<0  ⇒  ac>bca < b \;\text{and}\; c < 0 \;\Rightarrow\; ac > bc

Quadratic inequalities with a sign chart

Sign of a quadratic with a > 0 and roots x₁ < x₂

a(x−x1)(x−x2)<0  for  x1<x<x2,>0  for  x<x1  or  x>x2a(x - x_1)(x - x_2) < 0 \;\text{for}\; x_1 < x < x_2, \qquad > 0 \;\text{for}\; x < x_1 \;\text{or}\; x > x_2

Inequalities with absolute value

Absolute value inequalities (k > 0)

∣A∣<k  ⇔  −k<A<k∣A∣>k  ⇔  A<−k  or  A>k|A| < k \;\Leftrightarrow\; -k < A < k \qquad |A| > k \;\Leftrightarrow\; A < -k \;\text{or}\; A > k

Common traps

Dividing by a negative reverses the sign

From −3x<6-3x < 6 the answer is x>−2x > -2, not x<−2x < -2. The unreversed answer is always among the options.

x² < 16 is not x < 4

Taking the square root of both sides loses the negative half: x=−10x = -10 satisfies x<4x < 4 but not x2<16x^2 < 16. Write (x−4)(x+4)<0(x - 4)(x + 4) < 0 and read the answer −4<x<4-4 < x < 4 from the sign chart.

Do not multiply by a denominator of unknown sign

Multiplying 3x−1<1\frac{3}{x - 1} < 1 by x−1x - 1 assumes x−1>0x - 1 > 0 and loses every solution with x<1x < 1. Bring everything to one side and use a sign chart.

Greater than gives two pieces, not one

∣x+2∣>3|x + 2| > 3 is x<−5x < -5 or x>1x > 1. Writing it as −5>x>1-5 > x > 1 describes no number at all. Only the less-than form gives a single interval.

Equations with Fractions, Roots, Powers and Logs

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Equations with the unknown in a denominator

Clearing the denominator

P(x)Q(x)=R(x)  ⇒  P(x)=R(x) Q(x),Q(x)≠0\frac{P(x)}{Q(x)} = R(x) \;\Rightarrow\; P(x) = R(x)\,Q(x), \quad Q(x) \ne 0

Equations and inequalities with square roots

Squaring a root equation

A=B  ⇒  A=B2(keep only solutions with  B≥0)\sqrt{A} = B \;\Rightarrow\; A = B^2 \quad\text{(keep only solutions with}\; B \ge 0)

Exponential and logarithmic equations

Matching exponents or logs

af(x)=ag(x)  ⇒  f(x)=g(x)log⁡af(x)=log⁡ag(x)  ⇒  f(x)=g(x)>0a^{f(x)} = a^{g(x)} \;\Rightarrow\; f(x) = g(x) \qquad \log_a f(x) = \log_a g(x) \;\Rightarrow\; f(x) = g(x) > 0

Common traps

An excluded value can appear as an answer

After clearing denominators, the algebra may produce a value that makes an original denominator zero. It is not a solution. IMAT offers it as an option; the correct choice may be "no solution".

Squaring can add false solutions

Every solution of A=B\sqrt{A} = B solves A=B2A = B^2, but not the other way round. Substitute each answer back into the original equation; any that makes the right-hand side negative is false.

A root inequality has a domain

x<3\sqrt{x} < 3 is 0≤x<90 \le x < 9, not x<9x < 9: negative xx has no square root at all. Options that forget the lower limit are common.

Check the domain of every logarithm

Combining logs can produce an answer that makes an original argument zero or negative, such as x=−5x = -5 in log⁡x\log x. Such an answer is not a solution. Likewise, in a substitution t=axt = a^x, a negative tt gives no xx.

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