0580

Quadratic Equations

Algebra and Sequences

The rule for the first two terms

  • Completing the square rewrites a quadratic so that x appears only once, inside a squared bracket
  • Halve b and call the result p; the pair of terms x² + bx can then be replaced by (x + p)² − p²
    • The subtraction removes the extra p² that expanding the bracket creates
  • A negative b changes the sign inside the bracket but never the sign in front of p²
    • x² − 14x becomes (x − 7)² − 49
  • Once the first two terms are replaced, add on whatever constant the original had and tidy the numbers
  • Expanding your answer takes seconds and confirms it is right

When a number sits in front of x²

  • Remove it as a factor from the x² and x terms, leaving the constant sitting outside
  • Complete the square inside, multiply back through, and simplify until the expression reads a(x + p)² + q
  • Square brackets around the inner working keep it clearly separate from the squared bracket
Worked example

Completing the square with a leading number

Write 2x² − 12x + 5 in the form a(x + p)² + q.

Solution:

  • Completing the square rewrites the expression so x appears once, inside a squared bracket
  • The method needs the coefficient of x² to be 1, so take that factor out of the first two terms only, leaving the constant outside
  • 2x² − 12x + 5 = 2[x² − 6x] + 5
  • Inside the bracket, halve the coefficient of x: half of −6 is −3, so the bracket becomes (x − 3)²
  • But (x − 3)² expands to x² − 6x + 9, which is 9 too many, so subtract it back
  • x² − 6x = (x − 3)² − 9
  • Substitute that in: 2[(x − 3)² − 9] + 5
  • Multiply the 2 through the square brackets — it must reach the −9 as well: 2(x − 3)² − 18 + 5
  • Collect the constants: 2(x − 3)² − 13
  • Check by expanding: 2(x² − 6x + 9) − 13 = 2x² − 12x + 18 − 13 = 2x² − 12x + 5
  • Leaving the 2 outside only the bracket, as 2(x − 3)² − 9 + 5, is the standard slip

Reading off the turning point

  • A squared bracket can never be negative, so the smallest value it takes is zero, which happens when the bracket itself is zero
  • For y = a(x + p)² + q the curve turns at (−p, q), and the minus sign in the x-coordinate is the step most often missed
A negative quadratic drawn with its turning point marked by a cross, labelled to show that the completed square form a(x + p)² + q puts the turning point at (−p, q) whether the curve has a maximum or a minimum
Source: Completing the Square by Save My Exams
  • A positive a makes that point a minimum, and a negative a makes it a maximum
  • From 2(x − 3)² − 13 the bracket is zero at x = 3, so the turning point is (3, −13) and it is a minimum
  • The same "a square is never negative" argument is what proves a quadratic expression is always above some value
y = x² + 6x − 3 rewritten as y = (x + 3)² − 12: since a squared term is never smaller than zero, y is never smaller than −12
Source: Completing the Square by Save My Exams