Singapore-Cambridge A-Level Mathematics (H2)

Completing the square

Any quadratic $ax^2 + bx + c$ with $a \neq 0$ can be written as $a(x + p)^2 + q$. Taking out $a$ from the first two terms and halving the coefficient of $x$ gives

The square is never negative, so if $a > 0$ the least value of the expression is $c - \frac{b^2}{4a}$, reached when $x = -\frac{b}{2a}$. If $a < 0$ the same value is the greatest value. The turning point of the graph is therefore $\left(-\frac{b}{2a},\; c - \frac{b^2}{4a}\right)$ and the line $x = -\frac{b}{2a}$ is its axis of symmetry.

For $f(x) = 2x^2 - 12x + 7$:

The least value is $-11$, at $x = 3$, and the graph is a $\cup$-shaped curve with vertex $(3, -11)$.

Setting the completed form equal to zero and solving gives the quadratic formula,

The discriminant

The expression under the square root, $b^2 - 4ac$, is the discriminant, written $D$. It decides how many real roots $ax^2 + bx + c = 0$ has.

If $a > 0$ and $D < 0$, the graph lies wholly above the $x$-axis, so $ax^2 + bx + c > 0$ for every real $x$ (the expression is positive definite). If $a < 0$ and $D < 0$ the graph lies wholly below the axis, so the expression is negative for every real $x$.

Both conditions are needed. To show that $3x^2 - 4x + 5$ is always positive, note that $a = 3 > 0$ and $D = 16 - 60 = -44 < 0$.

Finding a range of values of a constant $k$ uses the discriminant as an inequality. For $x^2 + kx + (k + 3) = 0$ to have two distinct real roots,

so $k < -2$ or $k > 6$. For equal roots, $k = -2$ or $k = 6$.

A line meets a curve where their equations are solved together, and substituting the line into the curve gives a quadratic whose discriminant counts the intersections. If $D > 0$ the line cuts the curve at two points, if $D = 0$ it is a tangent, and if $D < 0$ it misses. For the line $y = x + c$ and the curve $y = x^2 - 3x + 5$, equating gives $x^2 - 4x + 5 - c = 0$, with $D = 16 - 4(5 - c) = 4c - 4$. The line is a tangent when $c = 1$.

Sum and product of roots

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, then

These follow from comparing $ax^2 + bx + c$ with $a(x - \alpha)(x - \beta)$. They give symmetric expressions in the roots without solving the equation, for example $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

For $2x^2 - 5x + 1 = 0$, $\alpha + \beta = \frac{5}{2}$ and $\alpha\beta = \frac{1}{2}$, so $\alpha^2 + \beta^2 = \frac{25}{4} - 1 = \frac{21}{4}$.

Simultaneous equations with one linear equation

When one equation is linear, solve it for one variable and substitute into the other. To solve $x + y = 5$ and $x^2 + xy - y^2 = 11$, write $y = 5 - x$:

So $x = 3, y = 2$ or $x = 12, y = -7$. Check the second in the first equation: $12 + (-7) = 5$, and in the second: $144 - 84 - 49 = 11$.

Surds

A surd is an irrational root such as $\sqrt{2}$ or $\sqrt[3]{5}$. The rules for square roots are

for positive $a$ and $b$. A surd is simplified by taking out square factors: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$.

Surds are added and subtracted only when the surd part is the same: $3\sqrt{12} - \sqrt{27} = 6\sqrt{3} - 3\sqrt{3} = 3\sqrt{3}$.

Brackets multiply as in ordinary algebra, using $(\sqrt{a})^2 = a$:

Rationalising the denominator

To remove a surd from a denominator, multiply top and bottom by something that makes the denominator rational. For a single surd, multiply by that surd. For a denominator of the form $a + \sqrt{b}$, multiply by its conjugate $a - \sqrt{b}$, because $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ is rational.

Equations involving surds

Isolate one surd, square both sides, and repeat if needed. Squaring can introduce solutions that do not satisfy the original equation, so each answer must be checked.

To solve $\sqrt{2x + 3} = x$:

Substituting $x = 3$ gives $\sqrt{9} = 3$, which is true. Substituting $x = -1$ gives $\sqrt{1} = -1$, which is false because a square root is non-negative. The only solution is $x = 3$.