Mike Cook A-Level Pure Maths Notes 1: Quadratics; Surds; Algebraic Division.

General Information

© M J Cook 2005. These notes can be reproduced for personal and educational use only. Commercial copying, hiring, and lending are prohibited. The material is loosely based on the AQA A-level pure maths syllabus and should be useful for students preparing for exams set by other UK exam boards. Current notes can be found at www.freewebs.com/mikecook.

Email: mikecook1982@yahoo.co.uk

Surds

A surd is a square root that cannot be expressed as a rational number or a quotient of real numbers. For example, (\sqrt{2} \approx 1.414213562…) is a surd.

Important Rules for Working with Surds:

  • (\sqrt{a} \times \sqrt{b} = \sqrt{ab})

  • (\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}})

  • (c\sqrt{a} \pm c\sqrt{b} = c(\sqrt{a} \pm \sqrt{b}))

Multiplying and Factorising Surds

The first rule can be used to multiply surds or to factorise surds, which can lead to simplifications.

Example: \sqrt{24} = \sqrt{3 \times 8} = \sqrt{8 \times 3}

Factorising Using Square Numbers

When faced with a surd, factorise using square numbers (4, 9, 16, 25, 36, 49, 64, etc.).

Example: \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}

Real Numbers as a Root

Any real number, like 3, 7, 49, etc., can be expressed as a root by remembering that (\sqrt{a^2} = a). For example, (\sqrt{5^2} = \sqrt{25} = 5). This allows multiplication of a surd by a real number by first expressing the real number as a root and then using the first rule of surds.

Example:

5\sqrt{12} - \sqrt{150} + \sqrt{2} = \sqrt{25} \sqrt{12} - \sqrt{150} + \sqrt{2} = \sqrt{25 \times 12} - \sqrt{150} + \sqrt{2} = \sqrt{300} - \sqrt{150} + \sqrt{2}

Rationalising the Denominator

It is usual to eliminate surds in the denominator where possible. For example, given (\frac{2}{\sqrt{3}}), eliminate the (\sqrt{3}) term in the denominator by multiplying numerator and denominator by (\sqrt{3}), i.e.,

\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

To rationalise the denominator of (\frac{2}{2+\sqrt{5}}), multiply the numerator and denominator by ({2-\sqrt{5}}). This technique uses the factorisation of a difference of squares. Recall: (a+b)(a-b) = a^2 - b^2.

Returning to our example, we have:

\frac{2}{2+\sqrt{5}} = \frac{2}{2+\sqrt{5}} \frac{2-\sqrt{5}}{2-\sqrt{5}} = \frac{2(2 - \sqrt{5})}{4 - 5} = -4 + 2\sqrt{5}

In general, to rationalise the denominator of (\frac{1}{a\sqrt{b} \pm c\sqrt{d}}), multiply numerator and denominator by (a\sqrt{b} \mp c\sqrt{d}).

General Process for Simplifying Surds

The general process for simplifying surds is to rationalise the denominator where appropriate, write the expression involving as few roots as possible, and write the roots as small as possible. Keep all rules and methods in mind.

Example: Simplify (\frac{3}{\sqrt{24}} + \sqrt{2} + \sqrt{6}). Start by rationalising the denominator.

(\frac{3}{\sqrt{24}} = \frac{3}{\sqrt{4 \times 6}} ) =\frac{3\sqrt{6}}{4}

Examples of Surd Simplification:

a) 5\sqrt{20} + 2\sqrt{45}

5\sqrt{4\times5} + 2\sqrt{9\times5} = 5 \times 2\sqrt{5} + 2 \times 3\sqrt{5} = 10\sqrt{5} + 6\sqrt{5} = 16\sqrt{5}

b) 3\sqrt{5} \times 2\sqrt{6} = 6\sqrt{30} \div \sqrt{12} = 6 \frac{\sqrt{30}}{\sqrt{4\times3}} = 6 \frac{\sqrt{30}}{2\sqrt{3}} = 3\sqrt{10}

Expressing in the Form a + b\sqrt{2}

Example: Express (\frac{2}{1 + \sqrt{2}}) in the form a + b\sqrt{2}, where a and b are integers.

The obvious thing to do here is to rationalise the denominator:

\frac{2}{1 + \sqrt{2}} = \frac{2}{1 + \sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{2(1-\sqrt{2})}{1-2} = -2 + 2\sqrt{2}

Quadratics

General Form

We are already familiar with the graphs of quadratic functions of the form f(x) = ax^2 + bx + c. We can also factorise quadratic equations of this form where possible to find the roots (zeros) or use the quadratic formula for finding the roots when factorisation is not possible.

Roots and Sketching

As a reminder, let us find the roots, and sketch, the following function: f(x) = x^2 - 14x + 3 . We know that the shape of this graph is a parabola (bucket shape). The parabola is the ‘right war round’ because the x^2 term is positive. Where does the parabola cross the x-axis? We call the point(s) where the parabola crosses the x-axis the roots (or zeros) of the quadratic.

Vertex and Line of Symmetry

Notice that the graph has a line of symmetry - a vertical line that cuts through the vertex. Of course, this property is seen in all quadratic parabolas. This is illustrated in fig1. 2. on the simplest quadratic parabola - f(x) = x^2 .

Key Methods for Finding Roots or Factors of Quadratic Polynomials:

  1. Write the quadratic in the form ax^2 + bx + c = 0. For example, write x^2=7x in the form x^2-7x = 0. Clear fractions where possible.

  2. If there is only one term involving x: Solve the equation by algebraic manipulation. For example, 2x^2=3.

  3. If there is one term involving x^2 and one term involving x and no constant terms - factorise out an x. For example, x = x^2.

  4. If the quadratic contains x^2 and x and constant terms, try to factorise into two linear factors. For example, x^2+4x - 5=0.

  5. If the quadratic does not factorise, use the quadratic formula. If ax^2 + bx + c = 0, then

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Finding Real Solutions

Example: Find the (real) solutions of the following: \frac{x^2+6}{x} = 12. This expression is a quadratic polynomial in disguise! Multiply throughout by x to clear the fraction:

x^2 - 12x + 6 = 0

Now factorise with care! We factorise to get:

(x - 5)(x-3) = 0

i.e. x = 5 or x = 3.

Example: Find the (real) solutions of the following: x^4 + 2x^2 - 63 = 0. Factorising out x^2 will not work, but we can substitute a = x^2 to turn this into a quadratic equation. Then:

(a + 9) (a - 7) = 0

x = \pm \sqrt{-9} , \pm \sqrt{7}

Completing the Square

Method

To complete the square of ax2 + bx + c = 0, we write,

(x + A)^2 + B = 0

To find A: Half the coefficient of x, i.e. A = \frac{1}{2}b

To find B: expand the bracket and see what needs to be added or subtracted to get equality.

Summary

SUMMARY: (ax^2 + bx + c = (x + A)^2 + B = 0),

where A = \frac{1}{2}b and B = c - \frac{b^2}{4}.

Applications and Useful Information

  • You can solve a quadratic equation.

  • Used to prove the quadratic equation formula

  • We can gain a useful piece of information by completing the square of a quadratic equation, that is we can say what the minimum value of the quadratic is.

Turning Points

The minimum (or maximum) point of a quadratic function is called the turning point of the function.

Example: Find the turning point (x, y) on the function y = x^2 + 16x + 63.

This function is a parabola (right way round), so the turning point is a minimum turning point. We can find this by completing the square:

(x + 8)^2 - 1 = 0

Since the minimum value of (x + 8)^2 = 0, the minimum value of y = x^2 + 16x + 63 occurs at y = −1. We find the x value at this point by substituting y = −1 into the equation and solving for x:

(x + 8)^2 - 1 = 0 can solve x = -9, -7

Discriminant

Any quadratic equation ax^2 + bx + c. can be categorised into cases depending on whether the roots are distinct, repeated or complex.

The information about whether the quadratic has two, one or no root(s) is contained in the term b^2 - 4ac .
This term is called the discriminant of the quadratic equation:

  • b^2 - 4ac > 0: Two distinct roots.

  • b^2 - 4ac = 0: One root (repeated).

  • b^2 - 4ac < 0: No real roots.

Algebraic Division

The process involves dividing a polynomial (quadratic or cubic) by a linear term. For example, dividing x^2 + 12x + 9 by x + 2.

General Form

Write f(x) = (x - a)(Quotient) + Remainder

  • (Quotient): The result of the division.

  • (Remainder): The value left after division.

We want to find linear term. Let linear term = ax + b write:

x2+= ax b+ + r(x) Remainder= a(b), is a constant, one degree less than the ‘linear term’.

Example:

(x^2 + 12x + 9) = (ax + b)(x + 2) + r

The expression on the LHS is the same as the expression on the RHS. From: ax \times x = x^2 it is clear that a = 1.

Conside the x term in both sides:
LHS = 12x.
RHS= 2 lot of a, b lots of x = 2x.
12x = 2a+b.
Since a =1 then, b = 10. rewrite equation:
(x^2 + 12x + 9) = (x + 10)(x + 2) + r$$