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:

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 quadratic and gives surds directly. Solving $x^2 - 6x + 2 = 0$ gives $(x - 3)^2 = 7$, so $x = 3 \pm \sqrt{7}$.

The quadratic formula comes from completing the square on $ax^2 + bx + c = 0$ in general:

For $3x^2 - 5x - 1 = 0$, $a = 3$, $b = -5$, $c = -1$, so $x = \frac{5 \pm \sqrt{25 + 12}}{6} = \frac{5 \pm \sqrt{37}}{6}$, which is $1.85$ or $-0.18$ to 2 decimal places.

The discriminant

The expression under the square root, $b^2 - 4ac$, is the discriminant. It decides how many real roots there are, without solving the equation.

A repeated root means the quadratic is a perfect square, $a(x + p)^2$, and the vertex lies on the $x$-axis.

Suppose $2x^2 + kx + 8 = 0$ is to have real roots. The condition is $k^2 - 64 \geq 0$, which means $(k - 8)(k + 8) \geq 0$, so $k \leq -8$ or $k \geq 8$. If the roots are to be equal then $k = \pm 8$ exactly, and if there are to be no real roots then $-8 < k < 8$.

The same test is used when a line meets a curve. Substituting the line into the curve's equation gives a quadratic, and its discriminant shows whether the line crosses the curve, touches it (a tangent) or misses it.

Quadratic inequalities

To solve a quadratic inequality, find the roots of the related equation and then sketch the graph. The sketch shows which side of the roots the expression is positive or negative, and this avoids sign mistakes.

Solve $x^2 - x - 6 < 0$. The roots of $(x - 3)(x + 2) = 0$ are $-2$ and $3$. The graph is a U-shape, so it is below the $x$-axis between the roots:

Solve $x^2 - 5x + 4 > 0$. The roots are $1$ and $4$, and the U-shape is above the axis outside them, so the solution is in two separate parts:

It is written with "or", because no number is both less than 1 and greater than 4. If the coefficient of $x^2$ is negative the graph is upside down, so sketch it rather than relying on the pattern above. For $-x^2 + 3x + 4 \geq 0$ the roots are $-1$ and $4$ and the graph is an upside-down U, so the solution is $-1 \leq x \leq 4$.

Simultaneous equations, one linear and one quadratic

Use the linear equation to write one unknown in terms of the other, then substitute into the quadratic. This gives a quadratic in one unknown.

Solve $x + y + 1 = 0$ and $x^2 + y^2 = 25$. From the first, $y = -x - 1$, so

so $x = -4$ gives $y = 3$, and $x = 3$ gives $y = -4$. The answers must be given as pairs, because each $x$ is matched to one $y$. Checking in the original equation, $16 + 9 = 25$ in both cases.

A second example: $2x + y = 5$ and $x^2 + xy + y^2 = 7$. Since $y = 5 - 2x$,

so $x = 2, y = 1$ or $x = 3, y = -1$.

The number of solutions is the number of intersections. A discriminant of zero after substitution means the line is a tangent, so there is one point of contact.

Equations that are quadratic in some function of $x$

An equation with powers $x^4$ and $x^2$, or $x$ and $\sqrt{x}$, or $\sin x$ and $\sin^2 x$, is a quadratic in disguise. Substitute a new letter for the repeated function, solve, and then go back to $x$.

For $x^4 - 13x^2 + 36 = 0$, let $u = x^2$. Then $u^2 - 13u + 36 = 0$, so $(u - 4)(u - 9) = 0$ and $u = 4$ or $9$. Therefore $x = \pm 2$ or $x = \pm 3$.

For $2x - 7\sqrt{x} + 3 = 0$, let $u = \sqrt{x}$, so $2u^2 - 7u + 3 = 0$ and $(2u - 1)(u - 3) = 0$. Then $u = \frac{1}{2}$ or $u = 3$, which gives $x = \frac{1}{4}$ or $x = 9$. Since $u = \sqrt{x}$ cannot be negative, any negative value of $u$ would have to be rejected.

For a trigonometric case, solve $2\cos^2 x + \cos x - 1 = 0$ for $0° \leq x \leq 360°$. With $c = \cos x$ this is $(2c - 1)(c + 1) = 0$. So $\cos x = \frac{1}{2}$, which gives $x = 60°$ or $300°$, and $\cos x = -1$, which gives $x = 180°$.

The limits on the new variable matter. If $u = \sin x$ then $-1 \leq u \leq 1$, and a root outside that interval gives no value of $x$.