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Linear and Quadratic Equations

An equation states that two expressions are equal.

534 xx

In solving an equation, operations are performed to isolate the variable or unknown.

This produces equivalent equations.

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Allowable operations include:

adding or subtracting the same quantity from each side of the equation

multiplying or dividing both sides of the equation by the same constant or some expression that includes the variable

simplifying or expanding either side of the equation to assist in its solution

raising both sides of an equation to the same power

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Example 1a

xx 1349 xxxx 13131349

044 x

40444 x44 x

4

4

4

4

x

1x

4

Example 1b

xx 1349 xxxx 913949

x44

4

4

4

4 x

x1

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Example 1c

xx 1349 xx 9134

x44 x1

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Example 2

4423 yy

4463 yy

yy 3446 y2

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Linear Equations

Linear equations can be written in the form

0bax

Why?

.0 constants, are and abaConditions:

Feature: the power of the variable, in this case, x, is always exactly 1.

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Often the equation will not be in the form

but can be manipulated and expressed in this way.

0bax

A good generalised model for solving equations is based on the process of

expansion isolation

consolidation

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Isolation

Isolation is the process of rearranging an equation in such a way that the unknown is left by itself on one side of an equation.

2 9 17x

2 9 9 17 9x

Take all the terms involving the unknown (just the 2x term) to the left and every other term to the right of the equals sign. We can do this by subtracting 9 from both sides:

2 8x

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Divide both sides by 2 to get the x by itself:

2 8x

2 82 2

4

x

x

Check that this is correct by substituting into the original equation:

2 4 9 17

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Consolidation

If the unknown appears more than once in an equation, consolidation is needed before proceeding to isolation.

5 9 2 9t t

Take all the terms involving the unknown to the left by subtracting 2t from both sides:

5 2 9 2 2 9

3 9 9

t t t t

t

12

3 9 9t

Now use isolation as before:

3 18t

183

t

6t

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Expansion

When the unknown appears in a bracket, expansion is needed.

Expand both sides:

5 7 2 1 27x x

5 35 2 2 27x x

and simplify

5 35 2 25x x

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Now consolidate:

and isolate:

5 2 35 2 2 25

3 35 25

x x x x

x

3 10

10 13

3 3

x

x

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How many solutions?

always true – true for all values of the unknown(s) sometimes true and sometimes false – true for some values of the unknown(s) never true – true for no value of the unknown(s)

An equation can be:

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Example

4 8 2 28 6 36x x x

Now attempting to consolidate:

8 8

which is true for all values of x.

4 2 2 14 6 6x x x

4 8 4 8x x

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Example

4 12 2 28 6 36x x x

Now attempting to consolidate:

12 8

Since this is clearly not true, the equation has no solution.

4 3 2 14 6 6x x x

4 12 4 8x x

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Further examples are given in the study guide. Pay particular attention to

Example 3-10 which shows what to do with messy or ugly equations.

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Equations Leading to Linear Equations

Consider the following equation

10)3)(2( 2 xxx

In this form, it does not appear to be linear at all but with a number of simple operations it can be expressed as follows and hence solved.

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10)3)(2( 2 xxx

10632 22 xxxx

1065 22 xxx

610522 xxx

165 x

5

16x

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Section 0.7 of the text also contains excellent examples of equations leading to linear equations.

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Of particular interest and providing the most challenge is the creation of the

mathematical equation from some verbal description of a process.

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A tank contains 25 litres of water but has the capacity for 1000 litres. Water begins pouring in at a rate of 15 litres per minute. How long will it take to fill the tank? Number of minutes = x

Example 4

15 25 1000x 1000 25

6515

x

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Consider a factory that produces moulded toys. It costs $3000 to set up a new mould and $5 for the materials to produce a single toy.

If the toys can be sold for $10 each, how many need to be produced to break even?

Linear Models

Cost/revenue model of manufacture

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The cost of manufacture can be represented by the expression 5x + 3000 where x is the number of toys manufactured.

This information is sometimes represented as the cost function and written as

C(x) represents the total cost of producing x toys.

5 3000C x x

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In general, C(x) = Bx+A is called the linear cost function.

A is the total fixed costs, that is, the sumof all costs not dependent upon the number of units manufactured (cost of machines, rental of floor space, insurance etc).

B is the total variable or per unit costs (materials used per unit, electricity cost per unit etc).

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In a similar way, the revenue generated by toy sales is given by 10x and incorporated in the revenue function as R(x)=10x.

In general, R(x)=Dx is called the linear revenue function where D is the unit selling price.

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By definition, the break-even point is the point where costs equal revenue, that is, C(x)=R(x).

xx 1030005

x600

x53000 xx 5103000

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A simple check shows that when x = 600, C(x) = R(x) = $6000.

It is easy to show that when x is less than 600, C(x) is greater than R(x), a loss . . .

and that when x is greater than 600, C(x) is less than R(x), a profit.

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Graphically, this can be shown as follows:

600

cost

revenue$

x

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The cost and revenue models are linear (can be graphed as straight lines), but in practice this may not be so.

To increase production past a certain point maymean working overtime so per unit costs are nolonger constant. Discounts for bulk orders may be given, etc. These situations require more elaborate models.

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Quadratic Equations

A quadratic equation can be written in the form

02 cbxax

0. and constants are and , acba

Feature: the highest power of the variable is 2.

Quadratic equations can be solved using factoring the quadratic formula

Conditions:

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Factoring

052 xx

Example 1

0)5( xx0x

5or x

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042 x

Example 2

0)2)(2( xx

202 xx

202or xx

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The Quadratic Formula

The solution to a quadratic equation is given by

a

acbbx

2

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Method:

1. Manipulate the equation into the standard form.

2. Identify the value of the constants a, b and c.

3. Substitute them into the formula and calculate.

There may be 0, 1 or 2 solutions. Why?

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2 solutions

1 solution

No solution

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Algebraically, the situation is described as follows:

is called the discriminant.acb 42

When the discriminant is less than 0, there are no solutions to the equation since the square root of a negative number is not real. When the discriminant equals 0, there is one solution and when the discriminant is greater than 0, two solutions can be found.

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Example 1

062 xx

1a 1b 6c

a

acbbx

2

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21 1 4 1 6

2 1x

2

2411 x

Quadratic Formula

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2

51x

2

51x

3x

2

5-1or

2or

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Example 2

0)2()1( 22 xx

04412 22 xxxx

0562 2 xx

5 6 2 cba

a

acbbx

2

42

41

26 6 4 2 52 2

4

40366

real.not is 4 sincesolution No

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Additional Features

As seen, the graph of quadratic functions arecurves called parabolas.

The curves will open upwards when a > 0 and have a minimum point and open downwards when a < 0 and have a maximum point.

The maximum or minimum point or turning point is called the vertex.

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It occurs when2

bx

a

2

b

a

0a

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Quadratic Models

If the toy manufacturer finds the price as a function of demand to be

find the equation for revenue and find the maximum revenue.

( ) 10p x x

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xxp )(R(x) = Price number of units

xx)10( 210 xx

xx 102 equation for revenue

Check: a = -1 < 0 so curve has maximum.

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Maximum occurs when

5

102( 1)

a

bx

2

Maximum revenue is achieved when 5 unitsare produced.

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If the costs of manufacture are $14 fixed and $1 per toy variable costs, determine the cost equation and any break-even points.

141)( xxC14x cost equation

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Break even points occur when

)()( xRxC

14 ,9 ,1 cba

xxx 1014 2 014102 xxx

01492 xx

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a

acbbx

2

42

29 9 4 1 14

2 1

2

59

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Costs = revenue generated when 2 or 7 units of toys are produced.

2

4or

2

14

2or 7

2

59

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Revenue

Cost

2 7

$