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Unit 1 Mathematical Methods

Chapter 3: Quadratics Objectives

• To recognise and sketch the graphs of quadratic polynomials. • To find the key features of the graph of a

quadratic polynomial: axis intercepts, turning point and axis of symmetry.

• To determine the maximum or minimum value of a quadratic function.

• To solve quadratic equations by factorising, completing the square and using the general formula.

• To apply the discriminant to determine the nature and number of solutions of a quadratic equation.

• To apply quadratic functions to solving problems.

What is a quadratic?

A quadratic is a Polynomial function follows the rule

𝒚𝒚 = 𝒂𝒂𝒏𝒏𝒙𝒙𝒏𝒏 + 𝒂𝒂𝒏𝒏−𝟏𝟏𝒙𝒙𝒏𝒏−𝟏𝟏 + …………. 𝒂𝒂𝟏𝟏𝒙𝒙 + 𝒂𝒂𝟎𝟎 𝒏𝒏 ∈ 𝑵𝑵

Where 𝑎𝑎0, 𝑎𝑎1, … … … … 𝑎𝑎𝑛𝑛 are co-efficients.

Degree of a polynomial is the highest power of x with a non-zero coefficient. 3A - Expanding and collecting like terms Example 1

Expand (2x – 3)(3√3 – 1) Use FOIL or other method Method 1 – FOIL = 2𝑥𝑥 × 3√3 – 2𝑥𝑥 × 1 − 3 × 3√3 + 3 × 1 = 6√3𝑥𝑥 - 2𝑥𝑥 − 9√3 + 3

Method 2 – other method = 2𝑥𝑥 (3√3 – 1) – 3 (3√3 – 1) = 6√3𝑥𝑥 - 2𝑥𝑥 − 9√3 + 3

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Menu → Algebra → Expand

Remember you need

to insert ‘x’ multiplication sign in between brackets

Perfect square ( 𝒂𝒂 + 𝒃𝒃 )𝟐𝟐 = 𝒂𝒂𝟐𝟐 + 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐

Example 1 Expand (𝒙𝒙 − √𝟓𝟓)𝟐𝟐 Recognise as a perfect square = 𝑥𝑥2 − 2√5 + (√5)2 = 𝑥𝑥2 − 2√5 + 5

Difference of two squares ( 𝒂𝒂 + 𝒃𝒃 )(𝒂𝒂 − 𝒃𝒃) = 𝒂𝒂𝟐𝟐 − 𝒃𝒃𝟐𝟐

Example 2 Simplify (𝟒𝟒𝒙𝒙 − 𝟕𝟕)(𝟒𝟒𝒙𝒙 + 𝟕𝟕) Recognise as a DOTS = (4𝑥𝑥)2 − 72 = 16x2 − 49

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Example 3 Find the area of the four rectangles.

𝑥𝑥 1 𝑐𝑐𝑐𝑐

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2

3

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Solution: 𝐴𝐴1 = 𝑥𝑥2 𝐴𝐴2 + 𝐴𝐴3 = 2(𝑥𝑥 × 1) 𝐴𝐴4 = 1 𝐴𝐴 = 𝑥𝑥2 + 2𝑥𝑥 + 1

Area of outer boundary ( 𝑥𝑥 + 1) × ( 𝑥𝑥 + 1)

𝐴𝐴= ( 𝑥𝑥 + 1 )2

𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵: 𝑥𝑥2 + 2𝑥𝑥 + 1 = ( 𝑥𝑥 + 1 )2 Perfect quare

Complete Exercise 3A Questions page 85

3B - Factorising

Steps to factorise

1. Always look for the highest common factor 2. Group terms 3. Recognise difference of two squares (DOTS) 4. Factorise quadratic expression

Example 1 Factorise 𝑥𝑥3 − 𝑏𝑏𝑥𝑥2 − 𝑎𝑎2𝑥𝑥 + 𝑎𝑎2𝑏𝑏 Solution: Group terms ( 𝑥𝑥3 − 𝑎𝑎2𝑥𝑥 ) + ( −𝑏𝑏𝑥𝑥2 + 𝑎𝑎2𝑏𝑏 ) 𝑥𝑥( 𝑥𝑥2 − 𝑎𝑎2) − 𝑏𝑏( 𝑥𝑥2 − 𝑎𝑎2 ) Common factor

( 𝑥𝑥2 − 𝑎𝑎2)( 𝑥𝑥 − 𝑏𝑏 )

DOTS ( 𝑥𝑥 − 𝑎𝑎 )( 𝑥𝑥 + 𝑎𝑎 )( 𝑥𝑥 − 𝑏𝑏 )

𝑥𝑥

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Example 2 Factorise 3𝑥𝑥2 + 15𝑥𝑥 + 18 = 3(𝑥𝑥2 + 5𝑥𝑥 + 6) = 3(𝑥𝑥 + 2 ) (𝑥𝑥 + 3) trinomial (factorise using cross method)

Factors of 3𝑥𝑥2 Factors of 18 ‘Cross-products’ add to give 15𝑥𝑥

3 𝑥𝑥

𝑥𝑥

+9 +2

+ 9 𝑥𝑥 + 6 𝑥𝑥 + 15 𝑥𝑥

= (3𝑥𝑥 + 9) (𝑥𝑥 + 2) take out common factor of 3 in first bracket = 3(𝑥𝑥 + 2 ) (𝑥𝑥 + 3)

Menu → Algebra → Factor

Remember you need to insert ‘x’ multiplication sign in between brackets

Complete Exercise 3B Questions page 91

3C - Quadratic equations

Steps to solve quadratic equations:

1. Write the equation in the form 𝒂𝒂𝒙𝒙𝟐𝟐 + 𝒃𝒃𝒙𝒙 + 𝒄𝒄 = 𝟎𝟎 2. Factorise the quadratic expression (ensure you equate

to zero) 3. Apply the null factor law to solve for the unknown.

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Example 1 Solve for 𝑥𝑥, in 10𝑥𝑥2 − 11𝑥𝑥 + 3 = 0

Factors of 10𝑥𝑥2

Factors of 3 ‘Cross-products’ add to give −11𝑥𝑥

5 𝑥𝑥

2𝑥𝑥

-3 -1

-6 𝑥𝑥 -5 𝑥𝑥 -11 𝑥𝑥

(5𝑥𝑥 − 3 ) (2𝑥𝑥 − 1) = 0 Hence, 𝑥𝑥 = 3

5 𝑜𝑜𝑜𝑜 𝑥𝑥 = 1

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Example 2

A polygon with 𝑛𝑛 sides has 𝑛𝑛(𝑛𝑛−3)

2 diagonals.

Find the number of sides a polygon with 65 diagonals has. Solution: 𝑛𝑛(𝑛𝑛 − 3)

2= 65

𝑛𝑛(𝑛𝑛 − 3) = 130 𝑛𝑛2 − 3𝑛𝑛 − 130 = 0 Use cross method to factorise: (𝑛𝑛 + 10)(𝑛𝑛 − 13) = 0

𝑛𝑛 = −10 and 𝑛𝑛 = 13

Reject since negative measure There fore 𝑛𝑛 = 13 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 - A polygon with 65 diagonals has 13 sides.

Complete Exercise 3C Questions page 95

3D - Graphing quadratics

Features of the graph of 𝒚𝒚 = 𝒙𝒙𝟐𝟐: ■ The graph is called a parabola. ■ The possible 𝑦𝑦-values are all positive real numbers and 0. (This is

called the range of the quadratic and is discussed in a more general context in Chapter 5.)

■ The graph is symmetrical about the 𝑦𝑦 − 𝑎𝑎𝑥𝑥𝑠𝑠𝑠𝑠. The line about which the graph is symmetrical is called the axis of symmetry.

■ The graph has a vertex or turning point at the origin (0,0). ■ The minimum value of 𝑦𝑦 is 0 and it occurs at the turning point.

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Transformations of 𝒚𝒚 = 𝒙𝒙𝟐𝟐 By a process called completing the square (re-visited in Section 3E), all quadratics in polynomial form 𝒚𝒚 = 𝒂𝒂𝒙𝒙𝟐𝟐 + 𝒃𝒃𝒙𝒙 + 𝒄𝒄 may be transposed into what will be called the turning point form:

𝒚𝒚 = 𝒂𝒂(𝒙𝒙 − 𝒉𝒉)𝟐𝟐 + 𝒌𝒌

Re-cap of transformations applied to 𝒚𝒚 = 𝒂𝒂(𝒙𝒙 − 𝒉𝒉)𝟐𝟐 + 𝒌𝒌

Transformation And

Equation

Graph Axis of symmetry&

Turning point

Effect

𝑎𝑎 = 2, ℎ = 0, 𝑘𝑘 = 0

𝑎𝑎 = −2, ℎ = 0, 𝑘𝑘 = 0

𝒚𝒚 = −𝟐𝟐𝒙𝒙𝟐𝟐

𝑥𝑥 = 0 TP: (0,0)

Reflection in the 𝑥𝑥 − 𝑎𝑎𝑥𝑥𝑠𝑠𝑠𝑠 (inverted parabola) Dilation of factor 2 from the 𝑥𝑥 −𝑎𝑎𝑥𝑥𝑠𝑠𝑠𝑠 (parabola moves closer to 𝑦𝑦 − 𝑎𝑎𝑥𝑥𝑠𝑠𝑠𝑠, thinner looking parabola)

𝑎𝑎 =12

, ℎ = 0, 𝑘𝑘 = 0

𝑎𝑎 = 1, ℎ = 2, 𝑘𝑘 = 0

𝑎𝑎 = 1,

ℎ = −2, 𝑘𝑘 = 0

𝒚𝒚 = (𝒙𝒙 + 𝟐𝟐)𝟐𝟐

𝑥𝑥 = −2 TP: (−2,0)

Translation of 2 units in the negative direction of/along the 𝑥𝑥 − 𝑎𝑎𝑥𝑥𝑠𝑠𝑠𝑠. (Moves the parabola 2 units to the left).

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𝑎𝑎 = 1, ℎ = 0, 𝑘𝑘 = 2

𝒚𝒚 = 𝒙𝒙𝟐𝟐 + 𝟐𝟐

𝑥𝑥 = 0 TP: (0,2)

Translation of 2 units in the positive direction of/along the 𝑦𝑦 −𝑎𝑎𝑥𝑥𝑠𝑠𝑠𝑠. (Moves the parabola 2 units up).

𝑎𝑎 = 1, ℎ = 0,

𝑘𝑘 = −2

𝑎𝑎 = −12

, ℎ = −2,

𝑘𝑘 = 2

𝑎𝑎 = 2, ℎ = 2,

𝑘𝑘 = −2

In Summary:

Complete Exercise 3D Questions page 101

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3E – Completing the suare and turning points Completing the square Example: Express each in the form 𝑦𝑦 = 𝑎𝑎(𝑥𝑥 − ℎ)2 + 𝑘𝑘, by completing the square. Hence state the corodinates of the turning point.

1. 𝒚𝒚 = 𝟔𝟔𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟏𝟏𝟐𝟐 Step1: factorise (if necessary) Step 2: complete the sqare Step 3: Tidy into required form Step 4: lose outer brackets Step 5: State TP

2. 𝒚𝒚 = 𝒙𝒙𝟐𝟐 − 𝟕𝟕𝒙𝒙 + 𝟐𝟐

3. Solve the equation −𝟑𝟑𝒙𝒙𝟐𝟐 − 𝟓𝟓𝒙𝒙 + 𝟐𝟐 = 𝟎𝟎 by first completing the square. Complete steps 1 to 4 as above. Step 5: Find solutions, equate to zero Step 6: Factorise using DOTS Step 7: State solutions

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The equation for the axis of symmetry of a parabola

Note: You do not need to replicate this proof. The 𝑥𝑥 −coordinate of the turning point is 𝒙𝒙 = − 𝒃𝒃

𝟐𝟐𝒂𝒂. Substitute this value into the quadratic

polynomial to find the 𝑦𝑦-coordinate of the turning point.

For a quadratic function written in polynomial form 𝑦𝑦 = 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 = 𝑐𝑐, the axis of symmetry of its graph has the equation 𝒙𝒙 = − 𝒃𝒃

𝟐𝟐𝒂𝒂.

Example:

From example 3 above 𝒚𝒚 = −𝟑𝟑𝒙𝒙𝟐𝟐 − 𝟓𝟓𝒙𝒙 + 𝟐𝟐, state the turning point using the axis of symmetry.

Complete Exercise 3E Questions page 105

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3F – Graphing quadratics in polynomial form It is not always essential to convert a quadratic to turning point form in order to sketch its graph. We can sometimes find the 𝑥𝑥 − and 𝑦𝑦 −axis intercepts and the axis of symmetry from polynomial form by other methods and use these details to sketch the graph.

Textbook examples:

1. Find the x- and y-axis intercepts and the turning point, and hence sketch the graph of 𝑦𝑦 = 𝑥𝑥2 − 4𝑥𝑥. Solution

2. Find the x- and y-axis intercepts and the turning point, and hence sketch the graph of 𝑦𝑦 = 𝑥𝑥2 − 9. Solution

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3. Find the 𝑥𝑥 − and 𝑦𝑦 −axis intercepts and the turning point, and hence sketch the graph of 𝑦𝑦 = 𝑥𝑥2 + 𝑥𝑥 − 12. Solution

Complete Exercise 3F Questions page 109

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3G – Solving quadratic inequalities Steps to solve quadratic inequalities :

Note: ‘Set of 𝑥𝑥 − values’, implies Domain. Example:

Solve 𝑥𝑥2 + 𝑥𝑥 − 12 > 0.

Complete Exercise 3G Questions page 111

3H – the general quadratic formula Not all quadratics can be factorised by inspection, and it is often difficult to find the x-axis intercepts this way. Therefore, we use a general formula for finding the solutions of a quadratic equation in polynomial form.

The solutions of the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, where 𝑎𝑎 ≠ 0, are given by the quadratic formula:

𝒙𝒙 =−𝒃𝒃 ± √𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄

𝟐𝟐𝒂𝒂

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Note: 𝒙𝒙 = − 𝒃𝒃𝟐𝟐𝒂𝒂

is the axis of symmetry. Also, from the formula it can be seen that:

■ If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 > 0, there are two solutions. ■ If , 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = 0there is one solution. ■ If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 < 0, there are no real solutions.

This will be covered in detail in Ex 3I. Example:

Solve 𝑦𝑦 = −3𝑥𝑥2 − 12𝑥𝑥 − 7 using the quadratic formula. Hence, sketch the graph and state the turning point and all intercepts.

Complete Exercise 3H Questions page 115

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3I – the discriminant

The expression under the square root sign is called the discriminant. We write 𝜟𝜟 = 𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄

In Summary:

Discriminant

Number of solutions (𝑥𝑥 − intercepts)

Corresponding graph Nature of solutions for the equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 where 𝑎𝑎, 𝑏𝑏 and 𝑐𝑐 rational numbers.

∆> 0

Two

If ∆ is a perfect square and ∆≠ 0, ⇒ the equation has two rational solutions. If ∆ is not a perfect square and ∆> 0, ⇒ the equation has two irrational solutions.

∆= 0

one

∆= 0, ⇒ the equation has one rational solution.

∆< 0

none

If ∆ is not a perfect square and ∆< 0, ⇒ the equation has no solutions.

Example:

1. Find the values of 𝑐𝑐 for which the equation 3𝑥𝑥2 − 2𝑐𝑐𝑥𝑥 + 3 = 0 has: a. One solution

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You will need to sketch the graph of the discriminant to answer parts b and c. b. No solutions

c. Two distinct solutions. Question 14, Exercise 3I 2. Find the discriminant of the equation 4𝑥𝑥2 + (𝑐𝑐 − 4)𝑥𝑥 − 𝑐𝑐 = 0, where 𝑐𝑐 is a rational number, and hence show that the equation has rational solution(s).

Complete Exercise 3I Questions page 119

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3J – Solving simultaneous linear and quadratic equations If we wish to find the point or points of intersection between a straight line and a parabola, we can solve the equations simultaneously. It should be noted that depending on whether the straight line intersects, touches or does not intersect the parabola we may get two, one or zero points of intersection. If there is one point of intersection between the parabola and the straight line, then the line is a tangent to the parabola. As we usually have the quadratic equation written with 𝑦𝑦 as the subject, it is necessary to have the linear equation written with 𝑦𝑦 as the subject. Then the linear expression for 𝑦𝑦 can be substituted into the quadratic equation. Example:

Question 2e, Exercise 3J 1. Solve the following pairs of simultaneous equations

𝑦𝑦 = 6 − 𝑥𝑥 − 𝑥𝑥2 𝑦𝑦 = −2𝑥𝑥 − 2

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Textbook example Prove that the straight line with equation 𝑦𝑦 = 1 − 𝑥𝑥 meets the parabola with the equation 𝑦𝑦 = 𝑥𝑥2 − 3𝑥𝑥 + 2 once only.

Note: The discriminant applied to the second equation, 𝑎𝑎𝑥𝑥2 + (𝑏𝑏 − 𝑐𝑐)𝑥𝑥 + (𝑐𝑐1 − 𝑐𝑐2) = 0, can be used to determine the number of intersection points.

■ If ∆> 0, there are two intersection points. ■ If ∆= 0, there is one intersection point. ■ If ∆< 0, there are no intersection points.

Complete Exercise 3J Questions page 122

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3K – Families of quadratic polynomial functions To find a quadratic rule to fit given points, first choose the best form of quadratic expression to work with. Then substitute in the coordinates of the known points to determine the unknown parameters. Some possible forms are given here:

One point is needed to determine 𝑎𝑎.

Two points are needed to determine 𝑎𝑎 and 𝑐𝑐.

Two points are needed to determine 𝑎𝑎 and 𝑏𝑏.

Three points are needed to determine 𝑎𝑎, 𝑏𝑏 and 𝑐𝑐.

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Textbook examples:

1. A parabola has 𝑥𝑥 −axis intercepts −3 and 4 and it passes through the point (1,24). Find the rule for this parabola. 2. A family of parabolas have rules of the form 𝑦𝑦 = 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 2, where 𝑎𝑎 ≠ 0.

a. For a parabola in this family with its turning point on the 𝑥𝑥-axis, find 𝑎𝑎 in terms of 𝑏𝑏.

b. If the turning point is at (4,0), find the values of 𝑎𝑎 and 𝑏𝑏. 3. A parabola passes through the points (1,4), (0,5) and (−1,10). Find the rule for this parabola.

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Read through other examples in textbook.

Complete Exercise 3K Questions page 129

3L – Quadratic models In this section it is shown how quadratics can be used to solve worded problems, including problems which involve finding the maximum or minimum value of a quadratic polynomial that has been used to model a ‘practical’ situation. Examples:

1. Jenny wishes to fence off a rectangular vegetable garden in her backyard. She has 20 m of fencing wire which she will use to fence three sides of the garden, with the existing timber fence forming the fourth side. Calculate the maximum area she can enclose.

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2. A cricket ball is thrown by a fielder. It leaves his hand at a height of 2 metres above the ground and the wicketkeeper takes the ball 60 metres away again at a height of 2 metres. It is known that after the ball has gone 25 metres it is 15metres above the ground. The path of the cricket ball is a parabola with equation 𝑦𝑦 = 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐.

a. Find the values of 𝑎𝑎, 𝑏𝑏 and 𝑐𝑐. b. Find the maximum height of the ball above the ground. c. Find the height of the ball when it is 5 metres horizontally before it hits the

wicketkeeper’s gloves.

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Complete Exercise 3L Questions page 134

Chapter 3 Checklist This is the minimum requirement, do more where you feel it is necessary.

Exercise 3A: Expanding and collecting like terms Q 3, 4, 5, 6, 7, 8, 9, 10-b Exercise 3B: Factorising Q 3, 4, 5, 6, 7, 8, 9, 10 Exercise 3C: Quadratic equations Q 1, 2, 4, 5, 6, 7, 10, 12 Exercise 3D: Graphing quadratics Q 1-d,f, 3 Exercise 3E: Completing the square & TP Q 3, 4, 5, 6 Exercise 3F: Graphing in polynomial form Q1, 2, 3, 4 Exercise 3G: Solving quadratic inequalities Q 1, 2-LHS, 3, 4, 5, 6, 7 Exercise 3H: The quadratic formula Q 1-a,d, 2-b,c, 3-RHS, 4-RHS Exercise 3I: The discriminant Q 1-b,e, 2-c,d,f, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 17 Exercise 3J: Solving simultaneous equations Q 1-a,c, 2-c,f, 3-b,d, 4, 5, 6, 7, 8, 9 Exercise 3K: Families of quadratic polynomial fns Q1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12, 13, 14, 16, 20, 21 Exercise 3L: Quadratic models Q1, 2, 3, 4, 5, 6, 7, 8, 9 Chapter 3 Review Questions All