Oxford MAT (Mathematics Admissions Test)

## The quadratic formula A **quadratic** is an expression $ax^2 + bx + c$ with $a \neq 0$. The equation $ax^2 + bx + c = 0$ has solutions $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$ The formula comes from completing the square (below), so it works for every quadratic, including those that do not factorise over the integers. Solve $2x^2 - 3x - 5 = 0$. Here $a = 2$, $b = -3$, $c = -5$, so $b^2 - 4ac = 9 + 40 = 49$ and $$ x = \frac{3 \pm 7}{4}, \quad\text{so } x = \tfrac{5}{2} \text{ or } x = -1. $$ Substituting $x = -1$ gives $2 + 3 - 5 = 0$, which checks the signs. ## Completing the square Writing $x^2 + px$ as $\left(x + \tfrac{p}{2}\right)^2 - \tfrac{p^2}{4}$ turns a quadratic into one squared bracket plus a constant. $$ x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7. $$ When the coefficient of $x^2$ is not 1, take it out of the first two terms first: $$ \begin{aligned} 3x^2 - 12x + 5 &= 3(x^2 - 4x) + 5 \\ &= 3\left[(x - 2)^2 - 4\right] + 5 \\ &= 3(x - 2)^2 - 7. \end{aligned} $$ A squared bracket is never negative, so $3(x-2)^2 - 7 \geq -7$ for every real $x$, with equality at $x = 2$. The minimum value of the quadratic is therefore $-7$, reached at $x = 2$. In general $a(x + h)^2 + k$ has its turning point at $(-h, k)$, which is a minimum if $a > 0$ and a maximum if $a < 0$. Applying the same steps to $ax^2 + bx + c$ gives $$ a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}, $$ so the axis of symmetry is $x = -\dfrac{b}{2a}$. Setting the expression to zero and…