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

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

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.

When the coefficient of $x^2$ is not 1, take it out of the first two terms first:

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

so the axis of symmetry is $x = -\dfrac{b}{2a}$. Setting the expression to zero and taking square roots gives the quadratic formula.

The discriminant

The quantity $\Delta = b^2 - 4ac$ sits under the square root in the formula, and its sign decides how many real roots there are.

A condition on $\Delta$ is a condition on a parameter. For $x^2 + kx + (k + 3) = 0$,

There are two distinct real roots when $k < -2$ or $k > 6$, one repeated root when $k = -2$ or $k = 6$, and none when $-2 < k < 6$. At $k = 6$ the equation is $x^2 + 6x + 9 = (x + 3)^2 = 0$, which confirms it.

If $\Delta < 0$ and $a > 0$ the quadratic is positive for every real $x$, and if $\Delta < 0$ and $a < 0$ it is negative for every real $x$. A statement such as "$x^2 + kx + 4 > 0$ for all $x$" therefore means $k^2 - 16 < 0$, that is $-4 < k < 4$.

Tangency and the discriminant

A line meets a parabola where the two equations are equal, which gives a quadratic in $x$. The line is a tangent exactly when that quadratic has a repeated root, so $\Delta = 0$.

Find $c$ so that $y = 4x + c$ touches $y = x^2 + 2x + 5$. Setting them equal gives $x^2 - 2x + (5 - c) = 0$, so

The repeated root is $x = 1$, where both curves have height 8, so the touching point is $(1, 8)$.

Sum and product of roots

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$ then $ax^2 + bx + c = a(x - \alpha)(x - \beta)$, and comparing coefficients gives

These let an expression in the roots be found without solving. For $x^2 - 5x + 3 = 0$,

They also build an equation from given roots: a quadratic with roots $\alpha + 1$ and $\beta + 1$ has sum $\alpha + \beta + 2 = 7$ and product $\alpha\beta + \alpha + \beta + 1 = 9$, so it is $x^2 - 7x + 9 = 0$.

Hidden quadratics

An equation that is not quadratic in $x$ may be quadratic in a function of $x$. Substituting that function, solving, and then reversing the substitution is the method.

• $x^4 - 5x^2 + 4 = 0$ is quadratic in $u = x^2$, giving $u^2 - 5u + 4 = 0$, so $u = 1$ or $4$ and $x = \pm 1$ or $\pm 2$.
• $x - 5\sqrt{x} + 6 = 0$ is quadratic in $u = \sqrt{x}$, giving $u = 2$ or $3$, so $x = 4$ or $9$. Since $u = \sqrt{x}$ must be non-negative, both are accepted.
• $2^{2x} - 3\cdot 2^x + 2 = 0$ is quadratic in $u = 2^x$, giving $u = 1$ or $2$, so $x = 0$ or $1$. A value $u \leq 0$ would have to be rejected, because $2^x$ is always positive.

The last step matters: each solution for $u$ has to be checked against what $u$ is allowed to be before it is turned back into $x$.