Revision Notes on Quadratics

In these notes I am going to address the first 4 specification points relating to quadratics, these are the

key ways in which we can express quadratics so that we can go on to solve them, sketch them or

comment on them:

Know how, if it is possible, to express a quadratic as a product of factors

Know how to complete the square of a quadratic

Know how to use this form to locate the vertex of the quadratic and its axis of symmetry.

“ Know how, if it is possible, to express a quadratic as a product of factors”

What does this actually mean? It’s quite simple, we need to be able to factorise quadratics so that we

can split them into two brackets, or factors. This is a GCSE topic! Once we have factorised an equation,

we can find its roots and either solve it , or sketch it.

Example :

x² - 13x + 40

(x - 8)(x - 5)

These numbers add together to make -13 , and multiply together to make 40

But what happens if the quadratic has a value before the x² term? In other words, the coefficient of x² is

not zero.

We might be able to make this easier to factorise by taking this number outside of a bracket.

2x²-26x + 80 = 0

2(x²-13x+20) = 0

The 2 is a common factor and can be taken out the bracket. Now you will recognise that inside the

bracket is the same as in the previous example. This can then be factorised in the same way.

2(x – 5)( x – 8) = 0

Although this may seem like cheating, it is accurate. As the function equals zero, we know that one of

the factors ( the brackets) must be equal to zero, as the only way you can multiply terms together and

get zero is if one of them equals zero. We know that 2 doesn’t equal zero, so we know that if we equate

the other two brackets to zero, we have the two roots of the equation.

From now on we need to remember that quadratics are often expressed as ax²+ bx + c

Sometimes there are not common factors between a, b and c, and so we cannot do this. We may have

to try different combinations of factors. The way to do this is to look at all the factors of the coefficient

of x2 and know that these are the starting point for the possible brackets.

So, for 6x2 + 11x + 4

Factors of six are 6, 1, 3 and 2.

So, possible first terms in the two brackets are:

(3x )( 2x ) and (6x )(x )

Then we need to look at the combinations of factors for 4. The factors of 4 are 4, 1, 2 and 2. So the

possible bracket “endings” are:

( + 4 ) ( +1) and ( +2)( +2)

We then can look at the different combinations and work out which ones multiply out to the correct

quadratic. We can see that the working combination here is:

(3x + 4)( 2x + 1)

“Know how to complete the square of a quadratic”

Completing the square is useful when an equation has no real roots ( does not cross x axis). We can

complete the square in order to find the vertex ( top of the n or bottom of the u) of the parabola.

Steps:

1. Put the quadratic in ax² + bx + c form.

2. Half the coefficient of x, and put it in a bracket with x. Square the bracket, and subtract the half of the

coefficient squared.

3. Simplify.

But what if there is a number before the x²? This makes it much harder to complete the square. We

must first divide through by a. This might leave us with some fractions!

Steps:

1. Divide through by a. Put this outside of a bracket containing x² and a/bx. Don’t bother dividing the

constant, leave that outside the bracket.

2. Take half the coefficient and put it in a bracket with x, and minus it squared.

3. Multiply back through by three, to get rid of the double bracket and only leave one bracket ( which is

squared)

4. Simplify.

“Know how to use this form to locate the vertex of the quadratic and its axis of

symmetry.”

The completed square form can be used to locate the vertex, the minimum point and the axis of

symmetry.

If y= (x + p)2 + q

Then the vertex is ( - p, q)

The minimum value is q

And the equation of the line of symmetry is: x = -p

Quadratics y = bx + c is a straight line

ax² + bx + c where a, b and c are constants, is a quadratic.

You can write the quadratic expression x² - 6x + 8 in a number of

ways:

Factor form – (x – 4) (x – 2) useful for solving the quadratic

equation: x² - 6x + 8 = 0.

Completed square form – (x + 3)² - 1 useful for locating the

vertex of a parabola, which is the graph of y = x² - 6x + 8.

If you factorise you can find where the graph meets the x axis.

y = (x – 4) (x – 2) and y = 0 then x – 4 = 0 or x – 2 = 0

So x = 2 or x = 4. graph meets the x axis and (2,0) and (4,0)

Difference of two squares

p² - q² = (p – q)(p – q)

when p=x and q=3

x² - 9 = (x – 3)(x + 3)

when p=2x and q=5

4x² - 25 = (2x – 5)(2x +5)

Completing the square (x+ ½ b) = x² + bx + ¼ b², so x² + bx = (x + ½b)² - ¼ b² Now we add ‘c’ to both sides. x² + bx + c = (x² + bx) + c = (x + ½ b)² - ¼ b² + c Example: x² + 10x + 32 in completed square form. x² + 10x + 32 = (x² + 10x) + 32 = ((x + 5)² – 25) + 32 = (x + 5)² + 7

Expanding Brackets

Brackets should be expanded in the following ways: For an expression of the form a(b + c), the expanded version is ab + ac, multiply the term outside the bracket by everything inside the bracket (e.g. 2x(x + 3) = 2x² + 6x ) For an expression of the form (a + b)(c + d), the expanded version is ac + ad + bc + bd, in other words everything in the first bracket should be multiplied by everything in the second.

Example

Expand (2x + 3)(x - 1): (2x + 3)(x - 1) = 2x^2 - 2x + 3x - 3 = 2x^2 + x - 3

Factorising

Factorising is the reverse of expanding brackets, so it is, for example, putting 2x^2 + x - 3 into the form (2x + 3)(x - 1). This is an important way of solving quadratic equations. The first step of factorising an expression is to 'take out' any common factors which the terms have. So if you were asked to factorise x^2 + x, since x goes into both terms, you would write x(x + 1).

Factorising Quadratics

There is no simple method of factorising a quadratic expression, but with a little practise it becomes easier. One systematic method, however, is as follows:

Example

Factorise 12y^2 - 20y + 3 = 12y^2 - 18y - 2y + 3

The first two terms, 12y^2 and -18y both divide by 6y, so take out this factor of 6y. 6y(2y - 3) - 2y + 3 Now, make the last two expressions look like the expression in the bracket: 6y(2y - 3) -1(2y - 3) The answer is (2y - 3)(6y - 1)

Example:

Factorise x^2 + 2x - 8 x^2 + 4x - 2x - 8 x(x + 4) - 2x - 8 x(x + 4) - 2(x + 4) (x + 4)(x - 2)

The Difference of Two Squares

If you are asked to factorise an expression which is one square number minus another, you can factorise it immediately. This is because a^2 - b^2 = (a + b)(a - b) .

Example:

Factorise 25 - x^2 = (5 + x)(5 - x)

Completing the square When f(x) = ax^2 + bx + c it is written as f(x) = (x + p)^2 + q this is in the completing the square form We use it to locate: the vertex (the maximum/minimum value of y) And its axis of symmetry f(x) = x^2 + bx + c a =1 Halve the coefficient of the x term and write as (x + 1/2b)^2 You will need to subtract 1/4b^2 from c to balance the expression Example: f(x) = x^2 + 6x + 12 (x + 3)^2 – 9 + 12 (x + 3)^2 + 3

If a > 1 f(x) = ax^2 + bx + c Factorise first by taking a as a common factor of first 2 terms f(x) = a(x^2 + b/a x) + c f(x) = a((x + 1/2b/a)^ 2 – 1/4b^2/a^2) + c f(x) = a (x + 1/2b/a) ^2 – 1/4b^2/a + c example: f(x) = 3x^2 + 8x + 25 3(x^2 + 8/3x) + 25 3((x + 8/6) ^2 – 64/36) + 25 3(x + 4/3) ^2 – 64/12 + 25 3(x + 4/3) ^2 + 19 2/3

How to locate Vertex and axis of symmetry When a quadratic function is written as f(x) = (x + p)^2 + q then the axis of symmetry is at x = -p (make bracket 0) and the vertex is at (-p, q) Example: f(x) = x^2 + 5x + 8 (x + 5/2)^2 – 25/4 + 8 (x + 5/2)^2 + 7/4 So axis of symmetry is x = -5/2 Vertex is (-5/2, 7/4)

Solving ax^2 + bx + c = 0 by completing the square Example: (x + 2)^2 – 6 = 0 (x + 2)^2 = 6 x + 2 = ± √6 x = – 2 ± √6

The discriminant

b^2 – 4ac = the discrimant

The discriminant tells you about the nature of the roots of a quadratic equation given that a, b and c are rational numbers. It quickly tells you the number of real roots or in other words the number of x-intercepts associated with a quadratic equation.

There are 3 situations:

b^2 – 4ac > 0 2 intercepts b^2 – 4ac = 0 1 intercept

Quadratic formula

b^2 – 4ac < 0 0 intercepts Simultaneous equations

A pair of "Simultaneous equations" is two equations which are both true at the same time. You have two equations which have two unknowns to be found.

Example

A man buys 3 fish and 2 chips for £2.80 A woman buys 1 fish and 4 chips for £2.60 How much are the fish and how much are the chips?

First we form the equations. Let fish be f and chips be c. We know that: 3f + 2c = 280 (1) f + 4c = 260 (2)

Elimination

This involves changing the two equations so that one can be added or subtracted from the other to leave us with an equation with only one unknown. We can change the equations by multiplying them through by a constant- as long as we multiply both sides of the equation by the same number it will remain true.

In our above example:

Doubling (1) gives: 6f + 4c = 560 (3)

Since equation (2) has a 4c in it, we can subtract this from the new equation (3) and the c's will all have disappeared: (3)-(2) gives 5f = 300 f = 60 Therefore the price of fish is 60p

So we can put f=60 in either of our original equations. Substitute this value into (1): 3(60) + 2c = 280 2c = 100 c = 50 Therefore the price of chips is 50p

Substitution

The method of substitution involves transforming one equation into x = (something) or y = (something) and then substituting this something into the other equation.

So,

Rearrange one of the original equations to isolate a variable. Rearranging (2): f = 260 - 4c Substitute this into the other equation: 3(260 - 4c) + 2c = 280 780 - 12c + 2c = 280 10c = 500 c = 50 Substitute this into one of the original equations to get f = 60

Maths topic 1- Quadratics...

Know that a basic quadratic looks like (ax²+bx+c)

Y= means the same as f(x)=

Writing an equation in factor form solves =0

Roots are where the graph crosses the x axis when y=0

Sketching a quadratic:

Plot the 1/2 roots and y intercept and sketch either a

1.Solving a quadratic by factorising:

To find the two x roots

How to factorise:

No constant term...take common x factor out bracket

Difference of 2 squares...write in (a+b)(a-b)

Sum of 2 squares...Never factorises!!!

Common numerical factor...Take it out the bracket and ignore

1. Equate to 0 2. Factorise

3. Solve to find roots for the x axis

4. Intercept on y axis = the constant

2.Solving a quadratic by Completing the square:

To find the vertex of the parabola (Max/ Min point)

To find the axis of symmetry

How to complete the square:

F(x)= ax²+ bx+ c...... f(x)=(x+ p)²+ q

If co-eff. of a = 1:

1. (x+ ½b )²

2. Constant – (½ b)²

3. Simplify

In function = In brackets

+ + = 2x positive

- + = 2x negative

-- or + - =Positive + negative

If co-eff. of a > 1:

1. Take a out as common factor

2. a ( x² + b/a x) + c

3. a ((x + ½ b/a)² - b²/4a²) +c

4. a (x + b/2a)²- b²/4a + c

5. Simplify

Using the completed square to locate the vertex and axis of symmetry:

Use the completed the square... f(x)= (x+p)² + q

The axis is at x=-p

The vertex is at (-p, q)

Solving ax²+ bx + c = 0 by completing the square:

1. Complete the square

2. Take constant across to other side

3. Remove brackets and ²(by square rooting)

4. After = write ±√

5. Either solve, or leave in surd form

3.Polynomials:

Positive integer powers of x

3x⁰

2x¹

4x² + 7x + 2

3x³ + 2x² + 5

Use two polynomial expressions p(x) and q(x)

Add, subtract, Multiply(using a giant 3 step smiley)

4.Quadratic formula:

Rearrangement of completing the square

√

We substitute values in to find two roots ±

The discriminant:

The bit in the root...b² - 4ac

If = < 0 it has no roots, graph lies above/ below axis... Use completing the square

If=0 it has one root, graph just touches the axis

If= >0 it has two roots, graph crosses the axis twice

5.Points of intersection of 2 graph:

Uses simultaneous equations to find the x and y co-ordinates where the two graphs touch

Use either elimination to subtract one from the other

Or...

Substitution of y into an equation or vice versa with x

The discriminant when substitution is complete, indicates the number of points of intersection

To solve:

1. Turn into a quadratic

2. Equate to 0

3. Take common factor out bracket

4. Factorise

5. Find x values

6. Find y values

Roots of Equations

Ewan Calder

Consider V to mean square root

The methods

1. Factorisation

2. Quadratic formula

3. Completing the square

4. Using the graph

• Use in the above order, although if there are no roots the the completing the square method needs to be used.

The discriminant • The discriminant = the part of the quadratic formula that

determines how many roots of the equation there are: b²4ac

• A root is where the line/graph crosses the x axis: In quadratic equations this can occur, 0, 1 or 2 times -

• If b²4ac<0 there are no roots.

• If b²4ac=0 there is one root.

• If b²4ac>0 there are 2 roots.

If there is one root to the equation, it is also the vertex.

Factorisation

• No constant term e.g. 4x²+2x

• Take out x if it is greater than 1 e.g. 2x(2x+1).

• If it equation is the difference of 2 squares it has no x term, e.g. 4x²-9 (a+b)(a-b) = a²-b²

• The sum of two squares will never factorise – another method must be used.

Work-through... continued

E.g. When a=1,

x²+5x+6 becomes (x+2)(x+3)

If a>1 6x²+11x+4 becomes 2(3x²-7x+4) which becomes 2(3x-4)(x-1)

Then to work out the root of the equation e.g. For (x+2)(x+3), the equation must be equated to 0 then x+2=0 and x+3=0 thus x=-2 or x=-3

These are the values for which the graph crosses the x axis.

The Quadratic Formula

• The formula =

• Is used when a quadratic expression is in the form ax² + bx + c = 0

• As with factorisation it gives you the points at which the graph crosses the x axis, in other words, where the equation and y = 0.

Work-through

E.g. x²-4x-7=0

a= 1

b= -4 X = (4±V(16-(4×1×-7)))/2

c=-7 X = (4 ± 2V11)/2

x = 2±V11

Completing the square

• Can be used to fine the minimum point (vertex of the curve) y and x coordinates.

• Can be used to locate the axis of symmetry of the curve.

• Can be used to fins the roots of the equation.

• When f(x) = ax²+bx+c it is written as f(x) = (x+p)+q

• This is converted by using (x+1/2b)²-1/4b²+c

Work-through

If a =1 e.g. f(x) = x²+4x+7 (x+2)²-4+7 which becomes (x+2)²+3 If a>1 or a negative value e.g. f(x) = 3x²+8x+25 3(x²+8/3x)+25 3((x+8/6)²-64/36)+25 3(x+4/3)²-64/36+25 3(x+4/3)²+19²/3

From this the vertex = (-4/3, 19²/3), the axis of symmetry= -4/3 and the y intercept as given by the original equation = 25. This is because x+4/3=0 therefore x = -4/3. In other words, when f(x) = (x+p)²+q the axis of symmetry is at x=-p and the vertex is at (-p,q).

Finding the roots...

When solving ax²+bx+c = 0 by completing the square, to find the roots this method is used:

e.g. (x+2)²-6=0

(x+2)=6

x+2 = ±V6

X=-2±V6

If these are drawn on the graph:

X X

(-2-V6,0) (-2+V6,0)