Algebraic Expressions

Algebraic Expressions are made up of Constants, Variables and Number Operations(Addition, Subtraction, Division and Multiplication).

The variables are shown with letters such as x, y, a, b, p, m, n, etc.

The terms in an algebraic expression are separated by a plus or a minus sign.

Examples:

• 2x + 3y

This is an algebraic expression with two terms which are 2x and 3y.

• 2x ( 3y )

This is an algebraic expression with only one term.

• (2x + 3y)(2x – 3y)

This is also only one term because it is two expressions in brackets multiplied together.

Addition and Subtraction

What you need to know

• We can add or subtract like terms.

• If the terms are like, we add or subtract the coefficients.

• Like terms have the same variables (letters) and the variables must have the same exponents.

• We cannot add or subtract unlike terms.

Examples:

• 3x + 5x = 8x

• –3a + 10a = 7a

• 6x²y + 3x − 10x²y = −4x²y + 3x

Multiplication and Division

What you need to know

• Positive Number × Positive Number = Positive Answer.

a × b = ab

• Positive Number × Negative Number = Negative Answer.

a × -b = -ab

• Negative Number × Negative Number = Positive Answer.

-a × -b = ab

• When Multiplying Fractions, Multiply Numerators by Numerators, and Denominators by Denominators.

$\frac{m}{n}$ × $\frac{b}{d}$ = $\frac{mb}{nd}$

• When Adding or Subtracting Fractions, Multiply the Denominators, make that the Denominator.

Then Multiply the Numerators of one Fraction with the Denominators of the other and Add or Subtract Everything.

$\frac{m}{n}$ + $\frac{b}{d}$ = $\frac{md + bn}{nd}$

Examples:

• 3x × 5y² = 15xy²

• 3x × -5y² = -15xy²

• -3x × -5y² = 15xy²

• $\frac{6x}{7y}$ × $\frac{3}{5z}$ = $\frac{18x}{35yz}$

• $\frac{6x}{8y}$ + $\frac{3}{12z}$ = $\frac{( 6x ) ( 12z ) + ( 3 ) ( 8y )}{( 8y ) ( 12z )}$

= $\frac{3xz + y}{4yz}$

The Distributive Law

• c(a + b) = c × a + c × b = ac + bc

• (x + y)(a + b) = ax + bx + ay + by

Examples:

• -3x(5x - 6y) = -15x² + 18xy

• (2x + y)(3x - 2y) = 6x² - 4xy + 3xy - 2y²

= 6x² - xy - 2y²

Factorising

Factorise an expression means writing it as a product of its factors.

Ways to factorise an expression

• Finding the Common Factor

9x² – 6xy² = 3x(3x – 2y²)

• Factorising by grouping in Pairs and then finding a Common Factor

3xy – 2x + 3y – 2 = 3xy + 3y – 2x – 2

= 3y(x + 1) – 2(x + 1)

= (x + 1)(3y – 2)

• Factorising a Difference of two Squares

16x² – y² = (4x – y)(4x + y)

• Factorising a Difference of two Cubes

8x³ – y³ = (2x – y)(4x² + 2xy + y²)

• Factorising a Sum of two Cubes

27a³ + 64b³ = (3a + 4b)(9a² – 12ab + 16b²)

• Factorising a Trinomial

9x² + 5x – 4 = (9x – 4)(x + 1)

Factorising a Trinomial

ax² + bx + c

Steps

• Multiply the Coefficient of x² by the Constant Value

a × c = ac

• Write down all the Products of ac

ac = p × q

ac = m × n

... etc

• Write the middle term bx in terms of the product of ac

bx = mx + nx

• Group the four terms

ax² + mx + nx + c

• Then Factorise by taking out a Common Factor

Examples:

Given 3x² + 11x + 6

Steps

• 3 × 6 = 18

Multiplying the Coefficient of x² by the Constant Value

• 18 = 18 × 1

= 9 × 2

= 6 × 3

Products of 18

• 11x = 9x + 2x

Writing the middle term 11x in terms of the product of 18

• 3x² + 9x + 2x + 6

Grouping the four terms

• 3x ( x + 3 ) + 2 ( x + 3 )

= ( x + 3 ) ( 3x + 2 )

Factorising by taking out a Common Factor

∴ 3x² + 11x + 6 = ( x + 3 ) ( 3x + 2 )

Given 4x² + 9x - 13

Steps

• 4 × 13 = 52

Multiplying the Coefficient of x² by the Constant Value

• 52 = 52 × 1

= 26 × 2

= 13 × 4

Products of 52

• 9x = 13x - 4x

Writing the middle term 9x in terms of the product of 52

• 4x² + 13x - 4x - 13

Grouping the four terms

• 4x ( x - 1 ) + 13 ( x - 1 )

= ( 4x + 13 )( x - 1 )

Factorising by taking out a Common Factor

∴ 4x² + 9x - 13 = ( 4x + 13 )( x - 1 )

Given 8x² - 18x + 9

Steps

• 8 × 9 = 72

Multiplying the Coefficient of x² by the Constant Value

• 72 = 72 × 1

= 36 × 2

= 24 × 3

= 18 × 4

= 12 × 6

= 9 × 8

Products of 72

• - 18x = - 12x - 6x

Writing the middle term - 18x in terms of the product of 72

• 8x² - 12x - 6x + 9

Grouping the four terms

• 4x ( 2x - 3 ) - 3 ( 2x - 3 )

= ( 4x - 3 )( 2x - 3 )

Factorising by taking out a Common Factor

∴ 8x² - 18x + 9 = ( 4x - 3 )( 2x - 3 )

Quadratic Equations

The Standard Form of a Quadratic Equation

ax² + bx + c = 0

For Example:

• x² + 5x + 6 = 0

a = 1

b = 5

c = 6

• 3x² - 7x - 12 = 0

a = 3

b = - 7

c = - 12

Solving a Quadratic Equation

Factorising

‘Solve a Quadratic Equation’ means find the unknown value(s) of x in a given quadratic equation.

The x-values in a Quadratic Equation are also called the roots of the equation when the equation is equal to zero.

To Solve for x

• Write the Quadratic Equation in a Standard Form

• Factorise

• When you have a product = 0, Set each Factor = 0

• Then solve for x

Examples:

Solve for x : x² - 7x = - 10

• x² - 7x + 10 = 0

Writing the equation in standard form

• ( x - 5 ) ( x - 2 ) = 0

Factorising the Trinomial

• ( x - 5 ) = 0 or ( x - 2 ) = 0

Setting each Factor = 0

∴ x = 5 or x = 2

Solving for x

Solve for x : x ( x + 3 ) = 0

• x ( x + 3 ) = 0

We have a product = 0

• x = 0 or ( x + 3 ) = 0

Setting each Factor = 0

∴ x = 0 or x = - 3

Solving for x

Solve for x : x ( 2x - 5 ) = 12

• 2x² - 5x - 12 = 0

Writing the equation in standard form

• ( 2x + 3 ) ( x - 4 ) = 0

Factorising the Trinomial

• ( 2x + 3 ) = 0 or ( x - 4 ) = 0

Setting each Factor = 0

∴ x = - $\frac{3}{2}$ or x = 4

Solving for x

Completing the Square

‘Completing the Square’ means writing

y = ax² + bx + c

in the form

y = a ( x + p )² + q

To do this we can follow the following steps:

• Take out a factor to make the coefficient of x² to be one

y = a [x² + $\frac{b}{a}$ x + $\frac{c}{a}$ ]

• Take half of the coefficient of x and square it. Then Add and Subtract this answer to keep the equation balanced

y = a [x² + $\frac{b}{a}$ x + ( $\frac{b}{2a}$ )² + $\frac{c}{a}$ - ( $\frac{b}{2a}$ )²]

• Complete the Square by Factorising

y = a [(x + $\frac{b}{2a}$ )² + $\frac{c}{a}$ - ( $\frac{b}{2a}$ )²]

y = a (x + $\frac{b}{2a}$ )² + c - $\frac{b 2}{4a}$

• We have now Completed the Square by writing

y = ax² + bx + c

in the form

y = a ( x + p )² + q

where

a ≠ 0

p = $\frac{b}{2a}$

q = c - $\frac{b 2}{4a}$

Examples:

Write

y = 3x² + 12x + 9

in the form

y = a ( x + p )² + q

• y = 3x² + 12x + 9

• y = 3 [ x² + 4x + 3 ]

Take out a factor to make the coefficient of x² to be 1

• y = 3 [ x² + 4x + 2² + 3 - 2² ]

Take half of the coefficient of x and square it. Then Add and Subtracting this answer

• y = 3 [ ( x + 2 )² - 1 ]

Completing the Square by Factorising

• y = 3 ( x + 2 )² - 3

We have now completed the square

∴ y = 3x² + 12x + 9 = 3 ( x + 2 )² - 3

with a = 3, p = 2, q = - 3

Solve for x by using the method of completing the square:

–3x² + 5x + 4 = 0

• –3x² + 5x + 4 = 0

• x² – $\frac{5}{3}$ x – $\frac{4}{3}$ = 0

Take out a factor to make the coefficient of x² to be 1

• x² – $\frac{5}{3}$ x + ( $\frac{5}{6}$ )² – $\frac{4}{3}$ – ( $\frac{5}{6}$ )² = 0

Take half of the coefficient of x and square it. Then Add and Subtract

• ( x – $\frac{5}{6}$ )² – $\frac{73}{36}$ = 0

Completing the Square by Factorising

• ( x – $\frac{5}{6}$ )² = $\frac{73}{36}$

• x – $\frac{5}{6}$ = ± $\sqrt[]{\frac{73}{36}}$

• x = $\frac{5}{6}$ ± $\frac{\sqrt[]{73}}{6}$ = $\frac{5 ± \sqrt[]{73}}{6}$

∴ x = $\frac{5 + \sqrt[]{73}}{6}$ or x = $\frac{5 – \sqrt[]{73}}{6}$

∴ x = $\frac{5 + \sqrt[]{73}}{6}$ or

x = $\frac{5 – \sqrt[]{73}}{6}$

Solving Quadratic Equations using the Formula

Some Quadratic Equations cannot be factorised, so to find their roots we use a Quadratic Formula

x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

To find the roots we substitute the value of a, b and c into the Quadratic Formula

Examples:

Solve for x: x² – 5x + 3 = 0

• x² – 5x + 3 = 0

There are no factors of 3 that can give us 5 when added, therefore we use the quadratic formula to solve this equation.

• x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

a = 1 , b = – 5 , c = 3

x = $\frac{– ( – 5 ) ± \sqrt[]{( – 5 ) 2 – 4 ( 1 ) ( 3 )}}{2 ( 1 )}$

= $\frac{5 ± \sqrt[]{25 – 12}}{2}$ = $\frac{5 ± \sqrt[]{13}}{2}$

∴ x = $\frac{5 + \sqrt[]{13}}{2}$ or x = $\frac{5 – \sqrt[]{13}}{2}$

∴ x = $\frac{5 + \sqrt[]{13}}{2}$ or

x = $\frac{5 – \sqrt[]{13}}{2}$

Solve for x (correct to two decimal places):

4x² – 8x = 7

• 4x² – 8x = 7

• 4x² – 8x - 7 = 0

Writing the equation in standard form

• x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

a = 4 , b = – 8 , c = – 7

x = $\frac{– ( – 8 ) ± \sqrt[]{( – 8 ) 2 – 4 ( 4 ) ( – 7 )}}{2 ( 4 )}$

= $\frac{8 ± \sqrt[]{64 + 112}}{8}$ = $\frac{8 ± \sqrt[]{176}}{8}$

x = $\frac{8 + \sqrt[]{176}}{8}$ or x = $\frac{8 – \sqrt[]{176}}{8}$

x = $\frac{8 + \sqrt[]{176}}{8}$ or

x = $\frac{8 – \sqrt[]{176}}{8}$

∴ x = 2.66 or x = – 0.66

Quadratic Inequalities

The standard form of a Quadratic Inequality is

ax² + bx + c > 0

ax² + bx + c < 0

ax² + bx + c ≥ 0

ax² + bx + c ≤ 0

Solving Quadratic Inequalities

• Write the inequality in standard form

• If the value of a < 0, multiply the equation by –1.

If we multiply the equation by a negative number, the inequality sign changes.

If the sign is > , it changes to < and vice versa. If it is ≥ it changes to ≤ and vice versa.

• Factorise the inequality if it is possible or use the quadratic formula to obtain the critical values( where the expression = 0 )

• Indicate the critical values on a number line

• Choose values on a number line that are less than, between and greater than the critical values.
Substitute them into the expression. Then indicate on the number line, for each valur, If the answer is positive, or negative.

• Then find the solution for the inequality.

Examples:

Solve for x: x² < 25

• x² < 25

• x² – 25 < 0

Writing the inequality in standard form

• ( x – 5 ) ( x + 5 ) < 0

Factorising

• x² – 25 = 0

at

x = 5 and x = – 5

Critical Values

• The expression x² – 25 = 0 at – 5 and 5, so we indicate this on the number line.

+0-0+

||||

|- 55|

We also choose values less than – 5, between – 5 and 5, and also greater than 5 then substitute them into x² – 25.

We also choose values less than - 5, between - 5 and 5, and also greater than 5 then substitute them into

x² - 25.

If the answer is positive, then we indicate + on the number line. If the answer is negative, we indicate – on the number line.

• We have to find values of x where

x² – 25 < 0.

So looking at the number line we see that

x² – 25 < 0

for all values between – 5 and 5.

∴ The solution is : – 5 < x < 5

Solve for x: – 2 x² – x + 3 ≤ 0

• – 2 x² – x + 3 ≤ 0

• 2 x² + x – 3 ≥ 0

If the value of a < 0, multiply the equation by – 1 and change the inequality sign.

• ( 2 x + 3 ) ( x – 1 ) ≥ 0

Factorising

• 2 x² + x – 3 = 0

at

x = – $\frac{3}{2}$ and x = 1

Critical Values

• The expression 2 x² + x – 3 = 0 at – $\frac{3}{2}$ and 1 , so we indicate this on the number line.

+0-0+

||||

\frac{3}{2}

We also choose values less than – $\frac{3}{2}$ , between – $\frac{3}{2}$ and 1, and also greater than 1 then substitute them into

2x² + x – 3

If the answer is positive, then we indicate + on the number line. If the answer is negative, we indicate – on the number line.

• We have to find values of x where

2 x² + x – 3 ≥ 0.

So looking at the number line we see that

2 x² + x – 3 ≥ 0

for all values of x ≤ – $\frac{3}{2}$ and values of x ≥ 1.

∴ The solution is : x ≤ – $\frac{3}{2}$ or x ≥ 1

Simulteneous Equations

Solving for x and y simultaneously given two equations.

Steps

• Always label the equations

Equation 1 ................................... 1

Equation 2 ................................... 2

• Use one of the equations to make one of the unknowns the subjectof the equation. ( eg, use equation 1 to get equation 3 )

Equation 3 ................................... 3

• Substitute 3 into 2 to get an equation with one unknown then solve for the unknown.

• Substitute the solved unknown into 3 to solve forthe other unknown.

Examples

Solve for x and y simultaneously:

y + 2x – 2 = 0 and 2x² + y² = 3yx

• y + 2x – 2 = 0 ................................... 1

2x² + y² = 3yx ................................... 2

• y + 2x – 2 = 0

∴ y = 2 – 2x ................................... 3

• Substitute 3 into 2

2x² + y² = 3yx

2x² + ( 2 – 2x )² = 3 ( 2 – 2x ) x

2x² + 4 – 8x + 4x² = 6x – 6x²

12x² – 14x + 4 = 0

∴ 6x² – 7x + 2 = 0

• Solve for x

6x² – 7x + 2 = 0

( 3x – 2 ) ( 2x – 1 ) = 0

( 3x – 2 ) = 0 or ( 2x – 1 ) = 0

∴ x = $\frac{2}{3}$ or x = $\frac{1}{2}$

• Substitute the values of x into 3 to get the values of y

y = 2 – 2x

For x = $\frac{2}{3}$

y = 2 – 2 ( $\frac{2}{3}$ )

∴ y = $\frac{2}{3}$

For x = $\frac{1}{2}$

y = 2 – 2 ( $\frac{1}{2}$ )

∴ y = 1

∴ ( x ; y ) = ( $\frac{2}{3}$ ; $\frac{2}{3}$ ) and ( x ; y ) = ( $\frac{1}{2}$ ; 1 )

∴ ( x ; y ) = ( $\frac{2}{3}$ ; $\frac{2}{3}$ ) and

( x ; y ) = ( $\frac{1}{2}$ ; 1 )

Solve for x and y simultaneously:

2x + y = 3 and x² + y + x = y²

• 2x + y = 3 ................................... 1

x² + y + x = y² ................................... 2

• 2x + y = 3

∴ y = 3 – 2x ................................... 3

• Substitute 3 into 2

x² + y + x = y²

x² + 3 – 2x + x = ( 3 – 2x )²

x² + 3 – 2x + x = 9 – 12x + 4x²

3x² – 11x + 6 = 0

• Solve for x

3x² – 11x + 6 = 0

( 3x – 2 ) ( x – 3 ) = 0

( 3x – 2 ) = 0 or ( x – 3 ) = 0

∴ x = $\frac{2}{3}$ or x = 3

• Substitute the values of x into 3 to get the values of y

y = 3 – 2x

For x = $\frac{2}{3}$

y = 3 – 2 ( $\frac{2}{3}$ )

∴ y = $\frac{5}{3}$

For x = 3

y = 3 – 2 ( 3 )

∴ y = – 3

∴ ( x ; y ) = ( $\frac{2}{3}$ ; $\frac{5}{3}$ ) and ( x ; y ) = ( 3 ; – 3 )

∴ ( x ; y ) = ( $\frac{2}{3}$ ; $\frac{5}{3}$ ) and

( x ; y ) = ( 3 ; – 3 )

Nature of Roots

The roots of any Quadratic Equation

ax² + bx + c = 0

can be found using the Quadratic Formula

x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

The roots of a quadratic equation are the x-values when the equation is zero.

When you are asked to ‘determine the nature of the roots of an equation’, you are NOT asked to solve the equation.

Finding the Nature of Roots

To find the nature of the roots of a quadratic equation look at the value of the discriminant ( denoted by Δ )

Δ = b² – 4ac

• If Δ < 0 : The roots are non-real.

• If Δ = 0 : There are two equal, real and rational roots.

• If Δ > 0 : There are two real roots which may be rational or irrational.

• If Δ is a perfect square, the roots are rational.

• If Δ is not a perfect square, then the roots are irrational.

The nature of the roots also tells us about the x-intercepts of the graph of a quadratic equation.

If Δ < 0

The roots are non-real.

There are no x-intercepts.

If Δ = 0

Roots are real and equal.

There is only one x-intercept and it is at the turning point of the graph.

If Δ > 0

There are two roots which are real and unequal.

If Δ is a perfect square, roots are rational.

If Δ is not a perfect square, the roots are irrational.

Examples

If x = 2 is a root of the equation

3x² – 5x – 2k = 0

determine the value of k.

• If x = 2 is a root of the equation, then we can substitute x = 2 into the equation and solve for k.

• 3x² – 5x – 2k = 0

• 3 ( 2 ) ² – 5 ( 2 ) – 2k = 0

• 12 – 10 – 2k = 0

• 2k = 2

∴ k = 1

Show that the roots of

x² – 2x – 7 = 0

are irrational, without solving the equation.

• Given x² – 2x – 7 = 0

a = 1, b = – 2, c = – 7

• Δ = b² – 4ac = ( – 2 )² – 4 ( 1 ) ( – 7 ) = 4 + 28 = 32

• Δ = b² – 4ac = ( – 2 )² – 4 ( 1 ) ( – 7 )

= 4 + 28 = 32

• Δ > 0 and not a perfect square

∴ The roots will be irrational.

Show that

x² + x + 1 = 0

has no real roots.

• Given x² + x + 1 = 0

a = 1, b = 1, c = 1

• Δ = b² – 4ac = ( 1 )² – 4 ( 1 ) ( 1 ) = 1 - 4 = - 3

• Δ = b² – 4ac = ( 1 )² – 4 ( 1 ) ( 1 )

= 1 - 4 = - 3

• Δ < 0

∴ There are no real roots.

Algebra

Algebraic Expressions

Examples:

Addition and Subtraction

What you need to know

Examples:

Multiplication and Division

What you need to know

a × b = ab

a × -b = -ab

-a × -b = ab

m n × b d = mb nd

m n + b d = md + bn nd

Examples:

The Distributive Law

• c(a + b) = c × a + c × b = ac + bc

• (x + y)(a + b) = ax + bx + ay + by

Examples:

Algebra

Home

Algebraic Expressions

QUIZ

Factorising

Tutorial

Factorising

Ways to factorise an expression

9x 2 – 6xy 2 = 3x(3x – 2y 2)

3xy – 2x + 3y – 2 = 3xy + 3y – 2x – 2 = 3y(x + 1) – 2(x + 1) = (x + 1)(3y – 2)

16x 2 – y 2 = (4x – y)(4x + y)

8x 3 – y 3 = (2x – y)(4x 2 + 2xy + y 2)

27a 3 + 64b 3 = (3a + 4b)(9a 2 – 12ab + 16b 2)

9x 2 + 5x – 4 = (9x – 4)(x + 1)

Factorising a Trinomial

ax2 + bx + c

Steps

a × c = ac

... etc

bx = mx + nx

ax2 + mx + nx + c

• Then Factorise by taking out a Common Factor

Examples:

• 3 × 6 = 18 Multiplying the Coefficient of x2 by the Constant Value

• 11x = 9x + 2x Writing the middle term 11x in terms of the product of 18

• 3x2 + 9x + 2x + 6 Grouping the four terms

• 3x ( x + 3 ) + 2 ( x + 3 )

= ( x + 3 ) ( 3x + 2 ) Factorising by taking out a Common Factor

∴ 3x2 + 11x + 6 = ( x + 3 ) ( 3x + 2 )

• 4 × 13 = 52 Multiplying the Coefficient of x2 by the Constant Value

• 9x = 13x - 4x Writing the middle term 9x in terms of the product of 52

• 4x2 + 13x - 4x - 13 Grouping the four terms

• 4x ( x - 1 ) + 13 ( x - 1 )

= ( 4x + 13 )( x - 1 ) Factorising by taking out a Common Factor

∴ 4x2 + 9x - 13 = ( 4x + 13 )( x - 1 )

• 8 × 9 = 72 Multiplying the Coefficient of x2 by the Constant Value

• - 18x = - 12x - 6x Writing the middle term - 18x in terms of the product of 72

• 8x2 - 12x - 6x + 9 Grouping the four terms

• 4x ( 2x - 3 ) - 3 ( 2x - 3 )

= ( 4x - 3 )( 2x - 3 ) Factorising by taking out a Common Factor

∴ 8x2 - 18x + 9 = ( 4x - 3 )( 2x - 3 )

Algebraic Expressions

Tutorial

Factorising

QUIZ

Quadratic Equations

Tutorial

Quadratic Equations

ax2 + bx + c = 0

• x2 + 5x + 6 = 0

• 3x2 - 7x - 12 = 0

Solving a Quadratic Equation

Factorising

Examples:

• x2 - 7x + 10 = 0 Writing the equation in standard form

• ( x - 5 ) ( x - 2 ) = 0 Factorising the Trinomial

• ( x - 5 ) = 0 or ( x - 2 ) = 0 Setting each Factor = 0

∴ x = 5 or x = 2 Solving for x

• x ( x + 3 ) = 0 We have a product = 0

• x = 0 or ( x + 3 ) = 0 Setting each Factor = 0

∴ x = 0 or x = - 3 Solving for x

• 2x2 - 5x - 12 = 0 Writing the equation in standard form

$\frac{m}{n}$ × $\frac{b}{d}$ = $\frac{mb}{nd}$

$\frac{m}{n}$ + $\frac{b}{d}$ = $\frac{md + bn}{nd}$

9x² – 6xy² = 3x(3x – 2y²)

3xy – 2x + 3y – 2 = 3xy + 3y – 2x – 2

= 3y(x + 1) – 2(x + 1)

= (x + 1)(3y – 2)

16x² – y² = (4x – y)(4x + y)

8x³ – y³ = (2x – y)(4x² + 2xy + y²)

27a³ + 64b³ = (3a + 4b)(9a² – 12ab + 16b²)

9x² + 5x – 4 = (9x – 4)(x + 1)

ax² + bx + c

ax² + mx + nx + c

• 3 × 6 = 18

Multiplying the Coefficient of x² by the Constant Value

• 11x = 9x + 2x

Writing the middle term 11x in terms of the product of 18

• 3x² + 9x + 2x + 6

Grouping the four terms

= ( x + 3 ) ( 3x + 2 )

Factorising by taking out a Common Factor

∴ 3x² + 11x + 6 = ( x + 3 ) ( 3x + 2 )

• 4 × 13 = 52

Multiplying the Coefficient of x² by the Constant Value

• 9x = 13x - 4x

Writing the middle term 9x in terms of the product of 52

• 4x² + 13x - 4x - 13

Grouping the four terms

= ( 4x + 13 )( x - 1 )

Factorising by taking out a Common Factor

∴ 4x² + 9x - 13 = ( 4x + 13 )( x - 1 )

• 8 × 9 = 72

Multiplying the Coefficient of x² by the Constant Value

• - 18x = - 12x - 6x

Writing the middle term - 18x in terms of the product of 72

• 8x² - 12x - 6x + 9

Grouping the four terms

= ( 4x - 3 )( 2x - 3 )

Factorising by taking out a Common Factor

∴ 8x² - 18x + 9 = ( 4x - 3 )( 2x - 3 )

ax² + bx + c = 0

• x² + 5x + 6 = 0

• 3x² - 7x - 12 = 0

• x² - 7x + 10 = 0

Writing the equation in standard form

• ( x - 5 ) ( x - 2 ) = 0

Factorising the Trinomial

• ( x - 5 ) = 0 or ( x - 2 ) = 0

Setting each Factor = 0

∴ x = 5 or x = 2

Solving for x

• x ( x + 3 ) = 0

We have a product = 0

• x = 0 or ( x + 3 ) = 0

Setting each Factor = 0

∴ x = 0 or x = - 3

Solving for x

• 2x² - 5x - 12 = 0

Writing the equation in standard form

• ( 2x + 3 ) ( x - 4 ) = 0

Factorising the Trinomial

• ( 2x + 3 ) = 0 or ( x - 4 ) = 0

Setting each Factor = 0

∴ x = - $\frac{3}{2}$ or x = 4

Solving for x

y = ax² + bx + c

y = a ( x + p )² + q

y = a [x² + $\frac{b}{a}$ x + $\frac{c}{a}$ ]

y = a [x² + $\frac{b}{a}$ x + ( $\frac{b}{2a}$ )² + $\frac{c}{a}$ - ( $\frac{b}{2a}$ )²]

y = a [(x + $\frac{b}{2a}$ )² + $\frac{c}{a}$ - ( $\frac{b}{2a}$ )²]

y = a (x + $\frac{b}{2a}$ )² + c - $\frac{b 2}{4a}$

• y = 3x² + 12x + 9

• y = 3 [ x² + 4x + 3 ]

Take out a factor to make the coefficient of x² to be 1

• y = 3 [ x² + 4x + 2² + 3 - 2² ]

Take half of the coefficient of x and square it. Then Add and Subtracting this answer

• y = 3 [ ( x + 2 )² - 1 ]

Completing the Square by Factorising

• y = 3 ( x + 2 )² - 3

We have now completed the square

• –3x² + 5x + 4 = 0

• x² – $\frac{5}{3}$ x – $\frac{4}{3}$ = 0

Take out a factor to make the coefficient of x² to be 1

• x² – $\frac{5}{3}$ x + ( $\frac{5}{6}$ )² – $\frac{4}{3}$ – ( $\frac{5}{6}$ )² = 0

Take half of the coefficient of x and square it. Then Add and Subtract

• ( x – $\frac{5}{6}$ )² – $\frac{73}{36}$ = 0

Completing the Square by Factorising

• ( x – $\frac{5}{6}$ )² = $\frac{73}{36}$

• x – $\frac{5}{6}$ = ± $\sqrt[]{\frac{73}{36}}$

• x = $\frac{5}{6}$ ± $\frac{\sqrt[]{73}}{6}$ = $\frac{5 ± \sqrt[]{73}}{6}$

∴ x = $\frac{5 + \sqrt[]{73}}{6}$ or x = $\frac{5 – \sqrt[]{73}}{6}$

∴ x = $\frac{5 + \sqrt[]{73}}{6}$ or x = $\frac{5 – \sqrt[]{73}}{6}$

∴ x = $\frac{5 + \sqrt[]{73}}{6}$ or

x = $\frac{5 – \sqrt[]{73}}{6}$

x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

• x² – 5x + 3 = 0

There are no factors of 3 that can give us 5 when added, therefore we use the quadratic formula to solve this equation.

• x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

a = 1 , b = – 5 , c = 3

x = $\frac{– ( – 5 ) ± \sqrt[]{( – 5 ) 2 – 4 ( 1 ) ( 3 )}}{2 ( 1 )}$

= $\frac{5 ± \sqrt[]{25 – 12}}{2}$ = $\frac{5 ± \sqrt[]{13}}{2}$

∴ x = $\frac{5 + \sqrt[]{13}}{2}$ or x = $\frac{5 – \sqrt[]{13}}{2}$

∴ x = $\frac{5 + \sqrt[]{13}}{2}$ or x = $\frac{5 – \sqrt[]{13}}{2}$

∴ x = $\frac{5 + \sqrt[]{13}}{2}$ or

x = $\frac{5 – \sqrt[]{13}}{2}$

• 4x² – 8x = 7

• 4x² – 8x - 7 = 0

Writing the equation in standard form

• x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

a = 4 , b = – 8 , c = – 7

x = $\frac{– ( – 8 ) ± \sqrt[]{( – 8 ) 2 – 4 ( 4 ) ( – 7 )}}{2 ( 4 )}$

= $\frac{8 ± \sqrt[]{64 + 112}}{8}$ = $\frac{8 ± \sqrt[]{176}}{8}$

x = $\frac{8 + \sqrt[]{176}}{8}$ or x = $\frac{8 – \sqrt[]{176}}{8}$

x = $\frac{8 + \sqrt[]{176}}{8}$ or x = $\frac{8 – \sqrt[]{176}}{8}$

x = $\frac{8 + \sqrt[]{176}}{8}$ or

x = $\frac{8 – \sqrt[]{176}}{8}$

ax² + bx + c > 0

ax² + bx + c < 0

ax² + bx + c ≥ 0

ax² + bx + c ≤ 0

• x² < 25

• x² – 25 < 0

Writing the inequality in standard form

• ( x – 5 ) ( x + 5 ) < 0

Factorising

• x² – 25 = 0

at

x = 5 and x = – 5

Critical Values