Cambridge International A-Level Mathematics

## Completing the square Any quadratic $ax^2 + bx + c$ with $a \neq 0$ can be written in the **completed square form** $a(x + p)^2 + q$. Expanding the bracket gives $ax^2 + 2apx + ap^2 + q$, so matching coefficients shows $p = \frac{b}{2a}$ and $q = c - ap^2$. For $2x^2 - 12x + 11$, take out the factor 2 from the terms in $x$ and complete the square inside the bracket: $$ \begin{aligned} 2x^2 - 12x + 11 &= 2(x^2 - 6x) + 11 \\ &= 2\left[(x - 3)^2 - 9\right] + 11 \\ &= 2(x - 3)^2 - 7 \end{aligned} $$ The form tells you the shape of the graph. A squared bracket is never negative, so $a(x + p)^2 + q$ has its least value $q$ when $a > 0$, reached at $x = -p$, and its greatest value $q$ when $a < 0$. The graph of $y = a(x + p)^2 + q$ has its vertex at $(-p, q)$ and the line $x = -p$ as its line of symmetry. In the example the vertex is $(3, -7)$, the minimum value of the expression is $-7$, and the graph crosses the $y$-axis at $(0, 11)$. The same idea finds the range of a function or proves that an expression is always positive. Since $x^2 - 4x + 7 = (x - 2)^2 + 3 \geq 3$, the expression is never zero, so the graph does not meet the $x$-axis. ## Solving quadratic equations A quadratic equation can be solved in three ways, and the choice depends on the numbers. Factorising is quickest when the roots are rational. For $x^2 + x - 12 = 0$ you need two numbers with product $-12$ and sum $1$, so $(x + 4)(x - 3) = 0$ and $x = -4$ or $x = 3$. Completing the square works for any…