Inequalities and Identities

Chapter 3 student notes

1. Simultaneous equations

Treat simultaneous equations as intersection problems; substitution turns geometry into algebra.

  1. Make one variable the subject when one equation is linear.
  2. Substitute into the nonlinear equation.
  3. Solve the resulting polynomial, normally a quadratic.
  4. Back-substitute for every valid value and present ordered pairs.

Linear + quadratic

\[y=2x-1,\qquad x^2+y^2=25.\] Substitution gives \[x^2+(2x-1)^2=25\Rightarrow 5x^2-4x-24=0.\] Solve for \(x\), then use \(y=2x-1\) for each root.
Exam tip. A simultaneous problem can have 0, 1, 2 or more intersections depending on the curves. Check that you report all solutions.

2. Linear, quadratic and rational inequalities

A sign diagram is more reliable than intuition, especially when factors or denominators repeat.

αβ++For a > 0, f(x) ≤ 0 exactly on α ≤ x ≤ β.
Linear sign rule
\[\text{multiplying or dividing by a negative reverses }<,>,\le,\ge\]
Quadratic sign idea
\[a(x-\alpha)(x-\beta)\gtrless0\quad\text{use root intervals}\]
  1. Move everything to one side.
  2. Factorise or find critical values.
  3. Place critical values on a number line.
  4. Test the sign in each interval or use the parabola shape.
  5. Include/exclude endpoints according to \(\le,\ge\) versus \(<,>\).

Worked example

Solve \(x^2-5x+6\le0\). \[(x-2)(x-3)\le0.\] The upward parabola is non-positive between its roots, so \[2\le x\le3.\]
Common mistake. Do not “take square roots” of a quadratic inequality as if it were an equation. The sign across intervals matters.

3. Polynomial division, factor theorem and remainder theorem

These theorems replace long division with one substitution when the divisor is linear.

Remainder theorem
\[\text{remainder on division by }(x-a)\text{ is }f(a)\]
Factor theorem
\[(x-a)\text{ is a factor}\iff f(a)=0\]

Find an unknown coefficient

If \(x-2\) is a factor of \(f(x)=x^3+kx^2-5x+6\), then \[f(2)=8+4k-10+6=0\Rightarrow4k+4=0\Rightarrow k=-1.\]

Remainder condition

If the remainder when \(f(x)\) is divided by \(x+3\) is 5, then use \(f(-3)=5\), not \(f(3)=5\).
Exam tip. When a cubic root is found, divide or factor it out immediately. The remaining quadratic often gives the rest of the solution.

4. Identities and coefficient comparison

Identities often appear as a setup for integration, partial rearrangement, proof or parameter determination.

Equating coefficients

If \[2x^2+7x+3\equiv A(x+1)^2+B(x+1)+C,\] expand the right-hand side, collect powers of \(x\), then equate the coefficients of \(x^2,x,1\). The symbol \(\equiv\) means the equality holds for every admissible \(x\).
Common mistake. An identity is not an equation to solve for \(x\). Its coefficients must match for all \(x\).

5. Chapter mastery checklist