Math Boot Camp - Class #3

Alex Vickery

Royal Holloway - University of London

26th September, 2017

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 1 / 35

Outline:Today’s Class

Equations:

Equations With Parameters

Quadratic Equations

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 2 / 35

Equations:

Outline:Today’s Class

Equations:

Equations With Parameters

Quadratic Equations

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 3 / 35

Equations:

Equations:Introduction:

Consider the following simple examples:

(a) 3x + 10 = x + 4, (b)z

z − 5+

1

3=−5

5− z, (c) Y = C + I

Equation (a) contains the variable x , whereas (b) has the variable z , andequation (c) has the three variables Y ,C and I .

To solve an equation means to find all values of the variables for whichthe equation is satisfied.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 4 / 35

Equations:

Equations:Introduction:

Consider the following simple examples:

(a) 3x + 10 = x + 4, (b)z

z − 5+

1

3=−5

5− z, (c) Y = C + I

Equation (a) contains the variable x , whereas (b) has the variable z , andequation (c) has the three variables Y ,C and I .

To solve an equation means to find all values of the variables for whichthe equation is satisfied.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 4 / 35

Equations:

Equations:Introduction:

Consider the following simple examples:

(a) 3x + 10 = x + 4, (b)z

z − 5+

1

3=−5

5− z, (c) Y = C + I

Equation (a) contains the variable x , whereas (b) has the variable z , andequation (c) has the three variables Y ,C and I .

To solve an equation means to find all values of the variables for whichthe equation is satisfied.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 4 / 35

Equations:

Equations:Introduction:

Consider the following simple examples:

(a) 3x + 10 = x + 4, (b)z

z − 5+

1

3=−5

5− z, (c) Y = C + I

Equation (a) contains the variable x , whereas (b) has the variable z , andequation (c) has the three variables Y ,C and I .

To solve an equation means to find all values of the variables for whichthe equation is satisfied.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 4 / 35

Equations:

Equations:Introduction:

For equation (a), this is easy.

In order to isolate the unknown x on one side of the equation, we add −xto both sides.

This gives 2x + 10 = 4

Adding -10 to both sides of the equation yields 2x = −6

Dividing by 2 we get the solution x = −3

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 5 / 35

Equations:

Equations:Introduction:

For equation (a), this is easy.

In order to isolate the unknown x on one side of the equation, we add −xto both sides.

This gives 2x + 10 = 4

Adding -10 to both sides of the equation yields 2x = −6

Dividing by 2 we get the solution x = −3

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 5 / 35

Equations:

Equations:Introduction:

For equation (a), this is easy.

In order to isolate the unknown x on one side of the equation, we add −xto both sides.

This gives 2x + 10 = 4

Adding -10 to both sides of the equation yields 2x = −6

Dividing by 2 we get the solution x = −3

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 5 / 35

Equations:

Equations:Introduction:

For equation (a), this is easy.

In order to isolate the unknown x on one side of the equation, we add −xto both sides.

This gives 2x + 10 = 4

Adding -10 to both sides of the equation yields 2x = −6

Dividing by 2 we get the solution x = −3

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 5 / 35

Equations:

Equations:Introduction:

For equation (a), this is easy.

In order to isolate the unknown x on one side of the equation, we add −xto both sides.

This gives 2x + 10 = 4

Adding -10 to both sides of the equation yields 2x = −6

Dividing by 2 we get the solution x = −3

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 5 / 35

Equations:

Equations:Introduction:

If any value of a variable makes an expression in a equation undefined,that value is not allowed.

Thus, the choice z = 5 is not allowed in (b).

This is because -5 makes the expressions zz−5 and −5

5−z undefined.

(Becoming 50 and −50 respectively).

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 6 / 35

Equations:

Equations:Introduction:

If any value of a variable makes an expression in a equation undefined,that value is not allowed.

Thus, the choice z = 5 is not allowed in (b).

This is because -5 makes the expressions zz−5 and −5

5−z undefined.

(Becoming 50 and −50 respectively).

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 6 / 35

Equations:

Equations:Introduction:

If any value of a variable makes an expression in a equation undefined,that value is not allowed.

Thus, the choice z = 5 is not allowed in (b).

This is because -5 makes the expressions zz−5 and −5

5−z undefined.

(Becoming 50 and −50 respectively).

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 6 / 35

Equations:

Equations:Introduction:

When faced with more complicated equations involving parentheses andfractions we follow these steps:

Begin by multiplying out the parentheses.

Multiply both sides of the equation by the lowest commondenominator for all the fractions.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 7 / 35

Equations:

Equations:Introduction:

When faced with more complicated equations involving parentheses andfractions we follow these steps:

Begin by multiplying out the parentheses.

Multiply both sides of the equation by the lowest commondenominator for all the fractions.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 7 / 35

Equations:

Equations:Introduction:

When faced with more complicated equations involving parentheses andfractions we follow these steps:

Begin by multiplying out the parentheses.

Multiply both sides of the equation by the lowest commondenominator for all the fractions.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 7 / 35

Equations:

Equations:Introduction:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 8 / 35

Equations:

Equations:Introduction:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 9 / 35

Equations:

Equations:Introduction:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 10 / 35

Equations With Parameters

Outline:Today’s Class

Equations:

Equations With Parameters

Quadratic Equations

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 11 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Simple relationships between two variables can often be described by alinear equation.

Examples of such linear equations in two variables are as follows:

(a) y = 10x , (b) y = 3x + 4, (c) y = −8

3x − 7

2

These equations have a common structure which makes it possible towrite down a general equation covering all the special cases:

y = ax + b

where a and b are real numbers.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 12 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Simple relationships between two variables can often be described by alinear equation.

Examples of such linear equations in two variables are as follows:

(a) y = 10x , (b) y = 3x + 4, (c) y = −8

3x − 7

2

These equations have a common structure which makes it possible towrite down a general equation covering all the special cases:

y = ax + b

where a and b are real numbers.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 12 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Simple relationships between two variables can often be described by alinear equation.

Examples of such linear equations in two variables are as follows:

(a) y = 10x , (b) y = 3x + 4, (c) y = −8

3x − 7

2

These equations have a common structure which makes it possible towrite down a general equation covering all the special cases:

y = ax + b

where a and b are real numbers.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 12 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Simple relationships between two variables can often be described by alinear equation.

Examples of such linear equations in two variables are as follows:

(a) y = 10x , (b) y = 3x + 4, (c) y = −8

3x − 7

2

These equations have a common structure which makes it possible towrite down a general equation covering all the special cases:

y = ax + b

where a and b are real numbers.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 12 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Simple relationships between two variables can often be described by alinear equation.

Examples of such linear equations in two variables are as follows:

(a) y = 10x , (b) y = 3x + 4, (c) y = −8

3x − 7

2

These equations have a common structure which makes it possible towrite down a general equation covering all the special cases:

y = ax + b

where a and b are real numbers.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 12 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Simple relationships between two variables can often be described by alinear equation.

Examples of such linear equations in two variables are as follows:

(a) y = 10x , (b) y = 3x + 4, (c) y = −8

3x − 7

2

These equations have a common structure which makes it possible towrite down a general equation covering all the special cases:

y = ax + b

where a and b are real numbers.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 12 / 35

Equations With Parameters

Equations With Parameters:Introduction:

The letters a and b are called parameters, and they take on differentvalues.

We often need to solve equations with “strange” letters denoting thevariables.

In addition, there may be several parameters involved.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 13 / 35

Equations With Parameters

Equations With Parameters:Introduction:

The letters a and b are called parameters, and they take on differentvalues.

We often need to solve equations with “strange” letters denoting thevariables.

In addition, there may be several parameters involved.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 13 / 35

Equations With Parameters

Equations With Parameters:Introduction:

The letters a and b are called parameters, and they take on differentvalues.

We often need to solve equations with “strange” letters denoting thevariables.

In addition, there may be several parameters involved.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 13 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Consider the basic macroeconomic model:

(i) Y = C + I , (ii) C = a + bY

Here, a and b are positive parameters of the model, with b < 1.

Equation (i) says that GDP, by definition, is the sum of consumption andinvestment.

Equation (ii) says that consumption is a linear function of GDP.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 14 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Consider the basic macroeconomic model:

(i) Y = C + I , (ii) C = a + bY

Here, a and b are positive parameters of the model, with b < 1.

Equation (i) says that GDP, by definition, is the sum of consumption andinvestment.

Equation (ii) says that consumption is a linear function of GDP.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 14 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Consider the basic macroeconomic model:

(i) Y = C + I , (ii) C = a + bY

Here, a and b are positive parameters of the model, with b < 1.

Equation (i) says that GDP, by definition, is the sum of consumption andinvestment.

Equation (ii) says that consumption is a linear function of GDP.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 14 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Consider the basic macroeconomic model:

(i) Y = C + I , (ii) C = a + bY

Here, a and b are positive parameters of the model, with b < 1.

Equation (i) says that GDP, by definition, is the sum of consumption andinvestment.

Equation (ii) says that consumption is a linear function of GDP.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 14 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Consider the basic macroeconomic model:

(i) Y = C + I , (ii) C = a + bY

Here, a and b are positive parameters of the model, with b < 1.

Equation (i) says that GDP, by definition, is the sum of consumption andinvestment.

Equation (ii) says that consumption is a linear function of GDP.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 14 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Solve the model for Y in terms of I and parameters:

Solution: Substituting C = a + bY into (i) gives:

Y = a + bY + I

Now rearrange this equation so that all the terms containing Y are on theleft hand side.

This can be done by adding −bY to both sides.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 15 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Solve the model for Y in terms of I and parameters:

Solution:

Substituting C = a + bY into (i) gives:

Y = a + bY + I

Now rearrange this equation so that all the terms containing Y are on theleft hand side.

This can be done by adding −bY to both sides.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 15 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Solve the model for Y in terms of I and parameters:

Solution: Substituting C = a + bY into (i) gives:

Y = a + bY + I

Now rearrange this equation so that all the terms containing Y are on theleft hand side.

This can be done by adding −bY to both sides.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 15 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Solve the model for Y in terms of I and parameters:

Solution: Substituting C = a + bY into (i) gives:

Y = a + bY + I

Now rearrange this equation so that all the terms containing Y are on theleft hand side.

This can be done by adding −bY to both sides.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 15 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Solve the model for Y in terms of I and parameters:

Solution: Substituting C = a + bY into (i) gives:

Y = a + bY + I

Now rearrange this equation so that all the terms containing Y are on theleft hand side.

This can be done by adding −bY to both sides.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 15 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Solve the model for Y in terms of I and parameters:

Solution: Substituting C = a + bY into (i) gives:

Y = a + bY + I

Now rearrange this equation so that all the terms containing Y are on theleft hand side.

This can be done by adding −bY to both sides.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 15 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Thus, cancelling the bY term on the right-hand side to give:

Y − bY = a + I

Notice that the left-hand side is equal to (1− b)Y , so (1− b)Y = a + I .

Dividing both sides by 1− b, then gives the answer which is:

Y =a

1− b+

1

1− bI

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 16 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Thus, cancelling the bY term on the right-hand side to give:

Y − bY = a + I

Notice that the left-hand side is equal to (1− b)Y , so (1− b)Y = a + I .

Dividing both sides by 1− b, then gives the answer which is:

Y =a

1− b+

1

1− bI

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 16 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Thus, cancelling the bY term on the right-hand side to give:

Y − bY = a + I

Notice that the left-hand side is equal to (1− b)Y , so (1− b)Y = a + I .

Dividing both sides by 1− b, then gives the answer which is:

Y =a

1− b+

1

1− bI

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 16 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Thus, cancelling the bY term on the right-hand side to give:

Y − bY = a + I

Notice that the left-hand side is equal to (1− b)Y , so (1− b)Y = a + I .

Dividing both sides by 1− b, then gives the answer which is:

Y =a

1− b+

1

1− bI

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 16 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Thus, cancelling the bY term on the right-hand side to give:

Y − bY = a + I

Notice that the left-hand side is equal to (1− b)Y , so (1− b)Y = a + I .

Dividing both sides by 1− b, then gives the answer which is:

Y =a

1− b+

1

1− bI

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 16 / 35

Equations With Parameters

Equations With Parameters:Introduction:

The solution is a formula expressing the endogeneous variable Y in termsof the exogenous variable I and the parameters a and b.

Note the power of this approach:

The model is solved only once, and then numerical answers are foundsimply by substituting appropriate numerical values for the parameters ofthe model.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 17 / 35

Equations With Parameters

Equations With Parameters:Introduction:

The solution is a formula expressing the endogeneous variable Y in termsof the exogenous variable I and the parameters a and b.

Note the power of this approach:

The model is solved only once, and then numerical answers are foundsimply by substituting appropriate numerical values for the parameters ofthe model.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 17 / 35

Equations With Parameters

Equations With Parameters:Introduction:

The solution is a formula expressing the endogeneous variable Y in termsof the exogenous variable I and the parameters a and b.

Note the power of this approach:

The model is solved only once, and then numerical answers are foundsimply by substituting appropriate numerical values for the parameters ofthe model.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 17 / 35

Equations With Parameters

Equations With Parameters:Introduction:

The solution is a formula expressing the endogeneous variable Y in termsof the exogenous variable I and the parameters a and b.

Note the power of this approach:

The model is solved only once, and then numerical answers are foundsimply by substituting appropriate numerical values for the parameters ofthe model.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 17 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Economists usually call the initial two equations, the structural form ofthe model.

Whereas the solution for Y is one part of the reduced form whichexpresses endogenous variables as functions of exogenous variables.

The other part of the reduced form is the equation:

C =(a + bI )

1− b

Which determines the second endogenous variable C .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 18 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Economists usually call the initial two equations, the structural form ofthe model.

Whereas the solution for Y is one part of the reduced form whichexpresses endogenous variables as functions of exogenous variables.

The other part of the reduced form is the equation:

C =(a + bI )

1− b

Which determines the second endogenous variable C .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 18 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Economists usually call the initial two equations, the structural form ofthe model.

Whereas the solution for Y is one part of the reduced form whichexpresses endogenous variables as functions of exogenous variables.

The other part of the reduced form is the equation:

C =(a + bI )

1− b

Which determines the second endogenous variable C .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 18 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Economists usually call the initial two equations, the structural form ofthe model.

Whereas the solution for Y is one part of the reduced form whichexpresses endogenous variables as functions of exogenous variables.

The other part of the reduced form is the equation:

C =(a + bI )

1− b

Which determines the second endogenous variable C .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 18 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Economists usually call the initial two equations, the structural form ofthe model.

Whereas the solution for Y is one part of the reduced form whichexpresses endogenous variables as functions of exogenous variables.

The other part of the reduced form is the equation:

C =(a + bI )

1− b

Which determines the second endogenous variable C .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 18 / 35

Equations With Parameters

Equations With Parameters:Introduction:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 19 / 35

Quadratic Equations

Outline:Today’s Class

Equations:

Equations With Parameters

Quadratic Equations

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 20 / 35

Quadratic Equations

Quadratic Equations:Introduction:

We now review the method for solving quadratic (also calledsecond-degree) equations.

The general quadratic equation has the form:

ax2 + bx + c = 0, (a 6= 0)

Where a, b and c are given constants, and x is the unknown.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 21 / 35

Quadratic Equations

Quadratic Equations:Introduction:

We now review the method for solving quadratic (also calledsecond-degree) equations.

The general quadratic equation has the form:

ax2 + bx + c = 0, (a 6= 0)

Where a, b and c are given constants, and x is the unknown.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 21 / 35

Quadratic Equations

Quadratic Equations:Introduction:

We now review the method for solving quadratic (also calledsecond-degree) equations.

The general quadratic equation has the form:

ax2 + bx + c = 0, (a 6= 0)

Where a, b and c are given constants, and x is the unknown.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 21 / 35

Quadratic Equations

Quadratic Equations:Introduction:

We now review the method for solving quadratic (also calledsecond-degree) equations.

The general quadratic equation has the form:

ax2 + bx + c = 0, (a 6= 0)

Where a, b and c are given constants, and x is the unknown.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 21 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If we divide each term by a, we get the equivalent equation:

x2 + (b/a)x + (c/a) = 0

If p = (b/a) and q = (c/a), the equation is:

x2 + px + q = 0

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 22 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If we divide each term by a, we get the equivalent equation:

x2 + (b/a)x + (c/a) = 0

If p = (b/a) and q = (c/a), the equation is:

x2 + px + q = 0

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 22 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If we divide each term by a, we get the equivalent equation:

x2 + (b/a)x + (c/a) = 0

If p = (b/a) and q = (c/a), the equation is:

x2 + px + q = 0

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 22 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If we divide each term by a, we get the equivalent equation:

x2 + (b/a)x + (c/a) = 0

If p = (b/a) and q = (c/a), the equation is:

x2 + px + q = 0

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 22 / 35

Quadratic Equations

Quadratic Equations:Introduction:

Two special cases are easy to handle.

If q = 0, (there is no constant term) the equation reduces to:

x2 + px = 0

This is equivalent to x(x + p) = 0.

Since the product of two numbers can only be zero if at least one of thenumbers is 0, we conclude that:

x = 0 or x = −p

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 23 / 35

Quadratic Equations

Quadratic Equations:Introduction:

Two special cases are easy to handle.

If q = 0, (there is no constant term) the equation reduces to:

x2 + px = 0

This is equivalent to x(x + p) = 0.

Since the product of two numbers can only be zero if at least one of thenumbers is 0, we conclude that:

x = 0 or x = −p

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 23 / 35

Quadratic Equations

Quadratic Equations:Introduction:

Two special cases are easy to handle.

If q = 0, (there is no constant term) the equation reduces to:

x2 + px = 0

This is equivalent to x(x + p) = 0.

Since the product of two numbers can only be zero if at least one of thenumbers is 0, we conclude that:

x = 0 or x = −p

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 23 / 35

Quadratic Equations

Quadratic Equations:Introduction:

Two special cases are easy to handle.

If q = 0, (there is no constant term) the equation reduces to:

x2 + px = 0

This is equivalent to x(x + p) = 0.

Since the product of two numbers can only be zero if at least one of thenumbers is 0, we conclude that:

x = 0 or x = −p

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 23 / 35

Quadratic Equations

Quadratic Equations:Introduction:

Two special cases are easy to handle.

If q = 0, (there is no constant term) the equation reduces to:

x2 + px = 0

This is equivalent to x(x + p) = 0.

Since the product of two numbers can only be zero if at least one of thenumbers is 0, we conclude that:

x = 0 or x = −p

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 23 / 35

Quadratic Equations

Quadratic Equations:Introduction:

Two special cases are easy to handle.

If q = 0, (there is no constant term) the equation reduces to:

x2 + px = 0

This is equivalent to x(x + p) = 0.

Since the product of two numbers can only be zero if at least one of thenumbers is 0, we conclude that:

x = 0 or x = −p

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 23 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If p = 0, (there is no term involving x).

Then the equation reduces to x2 + q = 0.

Then x2 = −q and there are two possibilities: If q > 0 then there are no

solutions. But if q ≤ 0, one has:

x2 + q = 0 ⇔ x = ±√−q (q ≤ 0)

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 24 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If p = 0, (there is no term involving x).

Then the equation reduces to x2 + q = 0.

Then x2 = −q and there are two possibilities: If q > 0 then there are no

solutions. But if q ≤ 0, one has:

x2 + q = 0 ⇔ x = ±√−q (q ≤ 0)

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 24 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If p = 0, (there is no term involving x).

Then the equation reduces to x2 + q = 0.

Then x2 = −q and there are two possibilities:

If q > 0 then there are no

solutions. But if q ≤ 0, one has:

x2 + q = 0 ⇔ x = ±√−q (q ≤ 0)

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 24 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If p = 0, (there is no term involving x).

Then the equation reduces to x2 + q = 0.

Then x2 = −q and there are two possibilities: If q > 0 then there are no

solutions. But if q ≤ 0, one has:

x2 + q = 0 ⇔ x = ±√−q (q ≤ 0)

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 24 / 35

Quadratic Equations

Quadratic Equations:Introduction:

If p = 0, (there is no term involving x).

Then the equation reduces to x2 + q = 0.

Then x2 = −q and there are two possibilities: If q > 0 then there are no

solutions. But if q ≤ 0, one has:

x2 + q = 0 ⇔ x = ±√−q (q ≤ 0)

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 24 / 35

Quadratic Equations

Quadratic Equations:Introduction:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 25 / 35

Quadratic Equations

Quadratic Equations:Harder Cases:

If both coefficients are different from 0, solving the equation becomesharder.

Consider, for example:

x2 − 4

3x − 1

4= 0

We could, of course, try to find values of x that solve the equation by trialand error.

However, it is not easy by this method to obtain the solutions which arex = 3

2 and x = −16 .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 26 / 35

Quadratic Equations

Quadratic Equations:Harder Cases:

If both coefficients are different from 0, solving the equation becomesharder.

Consider, for example:

x2 − 4

3x − 1

4= 0

We could, of course, try to find values of x that solve the equation by trialand error.

However, it is not easy by this method to obtain the solutions which arex = 3

2 and x = −16 .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 26 / 35

Quadratic Equations

Quadratic Equations:Harder Cases:

If both coefficients are different from 0, solving the equation becomesharder.

Consider, for example:

x2 − 4

3x − 1

4= 0

We could, of course, try to find values of x that solve the equation by trialand error.

However, it is not easy by this method to obtain the solutions which arex = 3

2 and x = −16 .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 26 / 35

Quadratic Equations

Quadratic Equations:Harder Cases:

If both coefficients are different from 0, solving the equation becomesharder.

Consider, for example:

x2 − 4

3x − 1

4= 0

We could, of course, try to find values of x that solve the equation by trialand error.

However, it is not easy by this method to obtain the solutions which arex = 3

2 and x = −16 .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 26 / 35

Quadratic Equations

Quadratic Equations:Harder Cases:

If both coefficients are different from 0, solving the equation becomesharder.

Consider, for example:

x2 − 4

3x − 1

4= 0

We could, of course, try to find values of x that solve the equation by trialand error.

However, it is not easy by this method to obtain the solutions which arex = 3

2 and x = −16 .

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 26 / 35

Quadratic Equations

Quadratic Equations:The General Case:

We now apply a method of completing the squares to the quadraticequation specified at the beginning.

This equation obviously has the same solutions as:

x2 + px = −q

One half of the coefficient of x is p/2.

Adding the square of this number to each side of the equation yields:

x2 + px + (p/2)2 = (p/2)2 − q

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 27 / 35

Quadratic Equations

Quadratic Equations:The General Case:

We now apply a method of completing the squares to the quadraticequation specified at the beginning.

This equation obviously has the same solutions as:

x2 + px = −q

One half of the coefficient of x is p/2.

Adding the square of this number to each side of the equation yields:

x2 + px + (p/2)2 = (p/2)2 − q

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 27 / 35

Quadratic Equations

Quadratic Equations:The General Case:

We now apply a method of completing the squares to the quadraticequation specified at the beginning.

This equation obviously has the same solutions as:

x2 + px = −q

One half of the coefficient of x is p/2.

Adding the square of this number to each side of the equation yields:

x2 + px + (p/2)2 = (p/2)2 − q

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 27 / 35

Quadratic Equations

Quadratic Equations:The General Case:

We now apply a method of completing the squares to the quadraticequation specified at the beginning.

This equation obviously has the same solutions as:

x2 + px = −q

One half of the coefficient of x is p/2.

Adding the square of this number to each side of the equation yields:

x2 + px + (p/2)2 = (p/2)2 − q

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 27 / 35

Quadratic Equations

Quadratic Equations:The General Case:

We now apply a method of completing the squares to the quadraticequation specified at the beginning.

This equation obviously has the same solutions as:

x2 + px = −q

One half of the coefficient of x is p/2.

Adding the square of this number to each side of the equation yields:

x2 + px + (p/2)2 = (p/2)2 − q

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 27 / 35

Quadratic Equations

Quadratic Equations:The General Case:

We now apply a method of completing the squares to the quadraticequation specified at the beginning.

This equation obviously has the same solutions as:

x2 + px = −q

One half of the coefficient of x is p/2.

Adding the square of this number to each side of the equation yields:

x2 + px + (p/2)2 = (p/2)2 − q

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 27 / 35

Quadratic Equations

Quadratic Equations:The General Case:

The left-hand side has now been made a complete square (of x+p/2), so:

(x + p/2)2 = p2/4− q

Note that if p2/4− q < 0, then the right-hand side is negative.

Since (x + p/2)2 is nonnegative for all choices of x .

We conclude that if p2/4− q < 0, then the equation has no solution.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 28 / 35

Quadratic Equations

Quadratic Equations:The General Case:

The left-hand side has now been made a complete square (of x+p/2), so:

(x + p/2)2 = p2/4− q

Note that if p2/4− q < 0, then the right-hand side is negative.

Since (x + p/2)2 is nonnegative for all choices of x .

We conclude that if p2/4− q < 0, then the equation has no solution.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 28 / 35

Quadratic Equations

Quadratic Equations:The General Case:

The left-hand side has now been made a complete square (of x+p/2), so:

(x + p/2)2 = p2/4− q

Note that if p2/4− q < 0, then the right-hand side is negative.

Since (x + p/2)2 is nonnegative for all choices of x .

We conclude that if p2/4− q < 0, then the equation has no solution.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 28 / 35

Quadratic Equations

Quadratic Equations:The General Case:

The left-hand side has now been made a complete square (of x+p/2), so:

(x + p/2)2 = p2/4− q

Note that if p2/4− q < 0, then the right-hand side is negative.

Since (x + p/2)2 is nonnegative for all choices of x .

We conclude that if p2/4− q < 0, then the equation has no solution.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 28 / 35

Quadratic Equations

Quadratic Equations:The General Case:

The left-hand side has now been made a complete square (of x+p/2), so:

(x + p/2)2 = p2/4− q

Note that if p2/4− q < 0, then the right-hand side is negative.

Since (x + p/2)2 is nonnegative for all choices of x .

We conclude that if p2/4− q < 0, then the equation has no solution.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 28 / 35

Quadratic Equations

Quadratic Equations:The General Case:

On the other hand, if p2/4− q > 0, then we have two possibilities:

x + p/2 =√

p2/4− q and x + p/2 = −√p2/4− q

Then the values of x are easily found.

These formulas are correct even if p2/4− q = 0

Though they give just the one solution x = −p/2 twice over.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 29 / 35

Quadratic Equations

Quadratic Equations:The General Case:

On the other hand, if p2/4− q > 0, then we have two possibilities:

x + p/2 =√

p2/4− q and x + p/2 = −√p2/4− q

Then the values of x are easily found.

These formulas are correct even if p2/4− q = 0

Though they give just the one solution x = −p/2 twice over.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 29 / 35

Quadratic Equations

Quadratic Equations:The General Case:

On the other hand, if p2/4− q > 0, then we have two possibilities:

x + p/2 =√

p2/4− q and x + p/2 = −√p2/4− q

Then the values of x are easily found.

These formulas are correct even if p2/4− q = 0

Though they give just the one solution x = −p/2 twice over.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 29 / 35

Quadratic Equations

Quadratic Equations:The General Case:

On the other hand, if p2/4− q > 0, then we have two possibilities:

x + p/2 =√

p2/4− q and x + p/2 = −√p2/4− q

Then the values of x are easily found.

These formulas are correct even if p2/4− q = 0

Though they give just the one solution x = −p/2 twice over.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 29 / 35

Quadratic Equations

Quadratic Equations:The General Case:

On the other hand, if p2/4− q > 0, then we have two possibilities:

x + p/2 =√

p2/4− q and x + p/2 = −√p2/4− q

Then the values of x are easily found.

These formulas are correct even if p2/4− q = 0

Though they give just the one solution x = −p/2 twice over.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 29 / 35

Quadratic Equations

Quadratic Equations:The General Case:

In Conclusion:

x2 + px + q = 0 if and only if x = −p2 ±

√p2

4 − q,

(p2

4 ≥ q

)

The Quadratic Formula:

ax2 + bx + c = 0 if and only if x = −b±√b2−4ac2a

If b2 − 4ac ≥ 0 and a 6= 0.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 30 / 35

Quadratic Equations

Quadratic Equations:The General Case:

In Conclusion:

x2 + px + q = 0 if and only if x = −p2 ±

√p2

4 − q,

(p2

4 ≥ q

)

The Quadratic Formula:

ax2 + bx + c = 0 if and only if x = −b±√b2−4ac2a

If b2 − 4ac ≥ 0 and a 6= 0.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 30 / 35

Quadratic Equations

Quadratic Equations:The General Case:

In Conclusion:

x2 + px + q = 0 if and only if x = −p2 ±

√p2

4 − q,

(p2

4 ≥ q

)

The Quadratic Formula:

ax2 + bx + c = 0 if and only if x = −b±√b2−4ac2a

If b2 − 4ac ≥ 0 and a 6= 0.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 30 / 35

Quadratic Equations

Quadratic Equations:The General Case:

In Conclusion:

x2 + px + q = 0 if and only if x = −p2 ±

√p2

4 − q,

(p2

4 ≥ q

)

The Quadratic Formula:

ax2 + bx + c = 0 if and only if x = −b±√b2−4ac2a

If b2 − 4ac ≥ 0 and a 6= 0.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 30 / 35

Quadratic Equations

Quadratic Equations:The General Case:

In Conclusion:

x2 + px + q = 0 if and only if x = −p2 ±

√p2

4 − q,

(p2

4 ≥ q

)

The Quadratic Formula:

ax2 + bx + c = 0 if and only if x = −b±√b2−4ac2a

If b2 − 4ac ≥ 0 and a 6= 0.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 30 / 35

Quadratic Equations

Quadratic Equations:The General Case:

It is a good idea to try and memorise this formula as once you have doneso, you can immediately write down the solutions to any quadraticequation.

Only if b−4ac ≥ 0 are the solutions real numbers.

If we use the formula when b2 − 4ac < 0, the square root of a negativenumber appears and no real solution exists.

The solutions are often called the roots of the equation.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 31 / 35

Quadratic Equations

Quadratic Equations:The General Case:

It is a good idea to try and memorise this formula as once you have doneso, you can immediately write down the solutions to any quadraticequation.

Only if b−4ac ≥ 0 are the solutions real numbers.

If we use the formula when b2 − 4ac < 0, the square root of a negativenumber appears and no real solution exists.

The solutions are often called the roots of the equation.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 31 / 35

Quadratic Equations

Quadratic Equations:The General Case:

It is a good idea to try and memorise this formula as once you have doneso, you can immediately write down the solutions to any quadraticequation.

Only if b−4ac ≥ 0 are the solutions real numbers.

If we use the formula when b2 − 4ac < 0, the square root of a negativenumber appears and no real solution exists.

The solutions are often called the roots of the equation.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 31 / 35

Quadratic Equations

Quadratic Equations:The General Case:

It is a good idea to try and memorise this formula as once you have doneso, you can immediately write down the solutions to any quadraticequation.

Only if b−4ac ≥ 0 are the solutions real numbers.

If we use the formula when b2 − 4ac < 0, the square root of a negativenumber appears and no real solution exists.

The solutions are often called the roots of the equation.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 31 / 35

Quadratic Equations

Quadratic Equations:The General Case:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 32 / 35

Quadratic Equations

Quadratic Equations:The General Case:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 33 / 35

Quadratic Equations

Quadratic Equations:The General Case:

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 34 / 35

Quadratic Equations

Quadratic Equations:The General Case:

If x1 and x2 are the roots of x2 + px + q = 0, then:

x1 + x2 = −p, and x1x2 = q

In words, the sum of the roots is minus the coefficient of the first-orderterm and the product is the constant term.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 35 / 35

Quadratic Equations

Quadratic Equations:The General Case:

If x1 and x2 are the roots of x2 + px + q = 0, then:

x1 + x2 = −p, and x1x2 = q

In words, the sum of the roots is minus the coefficient of the first-orderterm and the product is the constant term.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 35 / 35

Quadratic Equations

Quadratic Equations:The General Case:

If x1 and x2 are the roots of x2 + px + q = 0, then:

x1 + x2 = −p, and x1x2 = q

In words, the sum of the roots is minus the coefficient of the first-orderterm and the product is the constant term.

Alex Vickery (Royal Holloway) Math Boot Camp - Class #3 26th September, 2017 35 / 35