Completing the Square (AQA A-Level Mathematics): Revision Notes

Example: Expand the following complete square

$(x + 3)^2 = x^2 + 3x + 3x + 9 = x^2 + 6x + 9$

• Multiply $\frac{b}{2} \ by\ x$ and then square it.

Example: Write the following using a complete square

$x^2 + 6x$

• $(x + 3)^2$ gives us $x^2 + 6x + 9$
• This move has an extra $9$ that we do not want; we must subtract $9$ to make it equal to the line above. $(x + 3)^2 - 9$

Example: Write $x^2 + 12x$ in the form $(x + a)^2 + b$

$(x + 6)^2 - 36$

Example: Complete the square for $x^2 + 18x - 3$

$x^2 + 18x - 3 = (x + 9)^2 - 81 - 3 = (x + 9)^2 - 84$

• Complete the square for the underlined part:

$(x + 9)^2 - 84$

Example: Write $x^2 + 3x - 10$ in the form $(x + a)^2 + b$:

$x^2 + 3x - 10$

$= \left(x + \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 - 10$

$= \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} - 10$

$= \left(x + \frac{3}{2}\right)^2 - \frac{49}{4}$

• Much easier to work in fractions.

Example: Complete the square for $2x^2 + 12x - 1$

$2(x^2 + 6x) - 1$ Take out factor of $2$ so that we have something that says $x^2 + 6x$.

$2\left((x + 3)^2 - 9\right) - 1$ Ensure this is equal to the line above.

$2(x + 3)^2 - 18 - 1$ Multiply out square brackets.

$= 2(x + 3)^2 - 19$ You can multiply out to check it is equal to original answer.

Example: Write $-x^2 + 10x - 5$ in the form $a(x + b)^2 + c$:

$-x^2 + 10x - 5 \

= -[x^2 - 10x] - 5

\ = -[(x - 5)^2 - 25] - 5 \

= -(x - 5)^2 + 25 - 5

\ = -(x - 5)^2 + 20$

Example (Q4. Jun 2008, Q10) [Modified]: Express $2x^2 - 6x + 11$ in the form $p(x + q)^2 + r$:

$2(x^2 - 3x) + 11$

$= 2\left(x^2 - 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right) + 11$

$= 2\left((x - \frac{3}{2})^2 - \frac{9}{4}\right) + 11$

$= 2(x - \frac{3}{2})^2 - \frac{9}{2} + 11$

$= 2(x - \frac{3}{2})^2 + \frac{13}{2}$

Example: Find the coordinates of the vertex of $y = x^2 + 6x - 10$:

• Note: A vertex of a quadratic is its turning (max/min) point.

$y = (x + 3)^2 - 9 - 10 \Rightarrow y = (x + 3)^2 - 19$

$Y - coord$ is when $bracket$ $is$ $= 0$

$X- coord$ makes $bracket$ $0$

$x = -3 \quad y = -19$

$\therefore (-3, -19) \text{ is the vertex.}$

Example: Find the equation of the tangent to the following curve at its vertex and the corresponding line of symmetry. Use this to sketch the curve:

$y = x^2 - 3x + 4$

1. Completing the Square:

$y = (x - \frac{3}{2})^2 - \frac{9}{4} + 4$

$= (x - \frac{3}{2})^2 - \frac{9}{4} + \frac{16}{4}$

$= (x - \frac{3}{2})^2 + \frac{7}{4}$

• Vertex: $\left(\frac{3}{2}, \frac{7}{4}\right)$

Graph and Line of Symmetry:

• $V-shaped$, not $U-shaped$.
• Kinks and looks like the graph is about to go back down.
• Key points are not labelled.
• Line of symmetry: $x = \frac{3}{2}$
• Tangent: $y = \frac{7}{4}$

Q1 (Jun 2005, Q2): 2. Express $3x^2 + 12x + 7 in \ the\ form\ 3(x + a)^2 + b$:

$3(x^2 + 4x) + 7 \ = 3[(x + 2)^2-4]+7\\3(x+2)^2 -12 + 7 = 3(x + 2)^2 - 5$

3. Equation of the line of symmetry of the curve $y = 3x^2 + 12x + 7$:

• $x = -2$
• Key Points:
• $(x = -2)$ is the line of symmetry.
• The vertex is at $(x + 2)^2 - 5$.
• Illustration:
• Left: Tangent
• Right: Vertex