Completing the Square - Revision Notes

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Introduction

Completing the square is a method of rewriting a quadratic expression in the form $$(x + a)^2 + b$$ This technique helps in solving quadratic equations and is also useful in graphing parabolas.

Key Idea

The number inside the bracket is always **half the coefficient of $x$**.

Steps (when coefficient of $x^2 = 1$)

  1. Take half the coefficient of $x$.
  2. Square it and add/subtract appropriately to balance.
  3. Rewrite as a perfect square plus/minus a constant.

Example

Suppose we wish to complete the square for the quadratic expression \(x^2 + 5x + 3\).

This means we want to try to rewrite it so that it has the form of a complete square plus or minus a constant. The key point to remember is that the number in the bracket of the complete square is half the coefficient of \(x\).

So with \(x^2 + 5x + 3\) we know that the complete square will be \(\left(x + \tfrac{5}{2}\right)^2\). This has the same \(x^2\) and \(x\) terms as the given quadratic expression but the constant term is different. We must balance the constant term by subtracting the extra constant our complete square introduced \(\left(\tfrac{5}{2}\right)^2\), and adding the constant term from our quadratic, which is 3.

\[ x^2 + 5x + 3 = \left(x + \tfrac{5}{2}\right)^2 - \left(\tfrac{5}{2}\right)^2 + 3 \]

To finish off we just combine the two constants:

\[ -\left(\tfrac{5}{2}\right)^2 + 3 = -\tfrac{25}{4} + \tfrac{12}{4} = -\tfrac{13}{4} \]
\[ \therefore \quad x^2 + 5x + 3 = \left(x + \tfrac{5}{2}\right)^2 - \tfrac{13}{4} \]

We have now written the expression \(x^2 + 5x + 3\) as a complete square plus or minus a constant. Again note that the constant term, \(\tfrac{5}{2}\), inside the bracket is half the coefficient of \(x\) in the original expression.



Steps (when coefficient of $x^2 \neq 1$)

  1. Factor out the coefficient of $x^2$ from the quadratic terms.
  2. Complete the square inside the brackets.
  3. Simplify and expand if necessary.

Example

Suppose we wish to complete the square for the quadratic expression \(3x^2 - 9x + 50\).

We begin by factoring out the coefficient of \(x^2\), in this case 3. It does not matter that 3 is not a factor of 50; we can still do this by writing the expression as:

\[ 3x^2 - 9x + 50 = 3\left(x^2 - 3x + \tfrac{50}{3}\right) \]

Now the expression in brackets is a quadratic with coefficient of \(x^2 = 1\), so we can proceed as before. The number in the complete square will be half the coefficient of \(x\), so we use \(\left(x - \tfrac{3}{2}\right)^2\).

\[ x^2 - 3x + \tfrac{50}{3} = \left(x - \tfrac{3}{2}\right)^2 - \left(\tfrac{3}{2}\right)^2 + \tfrac{50}{3} \]

Now simplify the constants:

\[ -\left(\tfrac{3}{2}\right)^2 + \tfrac{50}{3} = -\tfrac{9}{4} + \tfrac{50}{3} = -\tfrac{27}{12} + \tfrac{200}{12} = \tfrac{173}{12} \]

So the quadratic inside the brackets becomes:

\[ x^2 - 3x + \tfrac{50}{3} = \left(x - \tfrac{3}{2}\right)^2 + \tfrac{173}{12} \]

Therefore, the original quadratic can be written as:

\[ 3x^2 - 9x + 50 = 3\left(\left(x - \tfrac{3}{2}\right)^2 + \tfrac{173}{12}\right) \]

Finally, expanding the factor of 3 across gives

\[ 3x^2 - 9x + 50 = 3\left(x - \tfrac{3}{2}\right)^2 + \tfrac{173}{4} \]

This is just a tidier version, with the constant simplified outside the brackets.



Summary

To complete the square for $ax^2 + bx + c$:
1. Factor out $a$ if $a \neq 1$.
2. Add and subtract $\left(\tfrac{b}{2a}\right)^2$.
3. Simplify into $(x + \tfrac{b}{2a})^2 + \text{constant}$.