# Quadratic Relations

### By: Harshpreet Singh

## Topics

- Expanding
- Factoring
- Solving
- Completing the Square

## Expanding

EXPANDING QUADRATIC EXPRESSIONS:

EXAMPLE ONE:

Expand ( x + 3 ) ( x + 2 ) without and with using FOIL.

**Answer (without using FOIL):**

( x + 3 ) ( x + 2 )

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

= x^2 + 2x + 3x + 6

= x^2 + 5x + 6

**Answer (with using FOIL):**

( x + 3 ) ( x + 2 )

= x^2 + 2x + 3x + 6

= x^2 + 5x + 6

EXAMPLE TWO:

( x + 4 ) ( x – 2 )

= x^2 – 2x + 4x – 8

= x^2 + 2x – 8

EXAMPLE THREE:

( 2x + 5 ) ( 3x – 8 )

= 6x^2 – 16x + 15x – 40

= 2x^2 – x – 40

QUESTIONS - EXPAND USING FOIL:

**Q1.** ( x + 6 ) ( x + 5 ) **Q2.** ( x – 5 ) ( x – 4 ) **Q3.** ( 2x + 5 ) ( 6x – 2 )

**Answers:** **A1.** x^2 + 11x + 30 **A2.** x^2 – 9x + 20 **A3.** 12x^2 + 26x – 10

PERFECT SQUARES:

( x + a )2 = x^2 + 2ax + a^2

( x – a )2 = x^2 – 2ax + a^2

EXAMPLE FOUR:

( x + 5 )2

= ( x + 5 ) ( x + 5 )

= x^2 + 10x + 25

EXAMPLE FIVE:

( x – 3 )2

= ( x – 3 ) ( x – 3 )

= x^2 – 6x + 9

QUESTIONS - EXPAND THESE PERFECT SQUARES:

**Q1.** ( x + 7 )^2 **Q2.** ( 2x + 5 )^2

**Answers:** **A1.** x^2 + 14x + 49 **A2.** 4x^2 + 20x + 25

DIFFERENCE OF SQUARES:

( x + a ) ( x – a ) = x^2 – a^2

EXAMPLE SIX:

( x + 5 ) ( x – 5 )

= x^2 – 5x + 5x – 25

= x^2 – 25

EXAMPLE SEVEN:

( x – 3 ) ( x + 3 )

= x^2 – 3x + 3x – 9

= x^2 – 9

QUESTIONS - EXPAND THESE DIFFERENCE OF SQUARES:

**Q1.** ( x + 7 ) ( x – 7 ) **Q2.** ( 2x + 5 ) ( 2x – 5 )

**Answers:** **A1.** x^2 – 49 **A2.** 4x^2 – 25

FACTORIZING QUADRATIC EXPRESSIONS:

Factorizing is the **reverse of expanding**.

EXAMPLE EIGHT:

x^2 + 6x + 5

= ( x + 5 ) ( x + 1 )

EXAMPLE NINE:

6x^2 + 2x – 20

= ( 2x + 4 ) ( 3x – 5 )

## Factoring

Factoring Quadratics

Ax^2 + bx + c = 0 **A Quadratic Equation in Standard Form**

(**a**, **b**, and **c** can have any value, except that **a** can't be 0.)

To "Factor" (or "Factorise" in the UK) a Quadratic is to:

find what to multiply to get the Quadratic

It is called "Factoring" because we find the factors (a factor is something we multiply by)

Example:

The factors of x^2 + 3x − 4 are:

(x+4) and (x−1)

**Why?** Well, let us multiply them to see:

(x+4)(x−1) = x(x−1) + 4(x−1)

= x^2 − x + 4x − 4

= x^2 + 3x − 4

Multiplying **(x+4)(x−1)** together is called Expanding.

In fact, Expanding and Factoring are opposites:

Expanding is easy, but Factoring can often be tricky

It is like trying to find out what ingredients

went into a cake to make it so delicious.

It can be hard to figure out!

So let us try an example where we **don't know** the factors yet:

Common Factor:

First check if there any common factors.

Example: what are the factors of 6x^2 − 2x = 0 ?

**6** and **2** have a common factor of **2**:

2(3x^2 − x) = 0

And **x^2** and **x** have a common factor of **x**:

2x(3x − 1) = 0

And we have done it! The factors are **2x** and **3x − 1**,

We can now also find the **roots** (where it equals zero):

- 2x is 0 when
**x = 0** - 3x − 1 is zero when
**x = 1/3**

And this is the graph (see how it is zero at x=0 and x=1/3):

But it is not always that easy ...

Guess and Check

Maybe we can guess an answer?

Example: what are the factors of 2x^2 + 7x + 3 ?

No common factors.

Let us try to **guess** an answer, and then check if we are right ... we might get lucky!

We could guess (2x+3)(x+1):

(2x+3)(x+1) = 2x^2 + 2x + 3x + 3

= 2x^2 + 5x + 3 **(WRONG)**

How about (2x+7)(x−1):

(2x+7)(x−1) = 2x^2 − 2x + 7x − 7

= 2x^2 + 5x − 7 **(WRONG AGAIN)**

OK, how about (2x+9)(x−1):

(2x+9)(x−1) = 2x^2 − 2x + 9x − 9

= 2x^2 + 7x − 9 **(WRONG AGAIN)**

Oh No! We could be guessing for a long time before we get lucky.

That is not a very good method. So let us try something else.

A Method For Simple Cases

Luckily there is a method that works in simple cases.

With the quadratic equation in this form:

Step 1: Find two numbers that multiply to give ac (in other words a times c), and add to give b.

Example: 2x^2 + 7x + 3

ac is 2×3 = **6** and b is **7**

So we want two numbers that multiply together to make 6, and add up to 7

In fact **6** and **1** do that (6×1=6, and 6+1=7)

How do we find 6 and 1?

It helps to list the factors of ac=**6**, and then try adding some to get b=**7**.

Factors of 6 include 1, 2, 3 and 6.

Aha! 1 and 6 add to 7, and 6×1=6.

Step 2: Rewrite the middle with those numbers:

Rewrite 7x with **6**x and **1**x:

2x^2 + **6x + x** + 3

Step 3: Factor the first two and last two terms separately:

The first two terms 2x^2 + 6x factor into 2x(x+3)

The last two terms x+3 don't actually change in this case

So we get:

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

Step 4: If we've done this correctly, our two new terms should have a clearly visible common factor.

In this case we can see that (x+3) is common to both terms

So we can now rewrite it like this:

2x(x+3) + (x+3) = (2x+1)(x+3)

Check: (2x+1)(x+3) = 2x^2 + 6x + x + 3 = **2x^2 + 7x + 3** (Yes)

Much better than guessing!

Let us try another example:

Example: 6x^2 + 5x − 6

**Step 1**: ac is 6×(−6) = **−36**, and b is **5**

List the positive factors of ac = **−36**: 1, 2, 3, 4, 6, 9, 12, 18, 36

One of the numbers has to be negative to make −36, so by playing with a few different numbers I find that −4 and 9 work nicely:

−4×9 = −36 and −4+9 = 5

**Step 2**: Rewrite **5x** with −4x and 9x:

6x^2 − 4x + 9x − 6

**Step 3**: Factor first two and last two:

2x(3x − 2) + 3(3x − 2)

**Step 4**: Common Factor is (3x − 2):

(2x+3)(3x − 2)

Check: (2x+3)(3x − 2) = 6x^2 − 4x + 9x − 6 = **6x^2 + 5x − 6** (Yes)

Finding Those Numbers:

The hardest part is finding two numbers that multiply to give ac, and add to give b.

It is partly guesswork, and it helps to **list out all the factors**.

Here is another example to help you:

Example: ac = −120 and b = 7

What two numbers **multiply to −120** and **add to 7** ?

The factors of 120 are (plus and minus):

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120

We can try pairs of factors (start near the middle!) and see if they add to 7:

- −10 x 12 = −120, and −10+12 = 2 (no)
- −8 x 15 = −120 and −8+15 = 7 (YES!)

Why Factor?

Well, one of the big benefits of factoring is that we can find the **roots** of the quadratic equation (where the equation is zero).

All we need to do (after factoring) is find where each of the two factors becomes zero

Example: what are the roots (zeros) of 6x^2 + 5x − 6 ?

We already know (from above) the factors are

(2x + 3)(3x − 2)

And we can figure out that

(2x + 3) is zero when x = −3/2

and

(3x − 2) is zero when x = 2/3

So the roots of 6x^2 + 5x − 6 are:

−3/2 and 2/3

Here is a plot of 6x^2 + 5x − 6, can you see where it equals zero?

And we can also check it using a bit of arithmetic:

At x = -3/2: 6(-3/2)^2 + 5(-3/2) - 6 = 6×(9/4) - 15/2 - 6 = 54/4 - 15/2 - 6 = 6-6 **= 0**

At x = 2/3: 6(2/3)^2 + 5(2/3) - 6 = 6×(4/9) + 10/3 - 6 = 24/9 + 10/3 - 6 = 6-6 **= 0**

Graphing:

We can also try graphing the quadratic equation. Seeing where it equals zero can give us clues.

Example: (continued)

Starting with 6x^2 + 5x − 6 and **just this plot:**

The roots are **around** x = −1.5 and x = +0.67, so we can **guess** the roots are:

−3/2 and 2/3

Which can help us work out the factors **2x + 3** and **3x − 2**

Always check though! 0.67 might not be 2/3 for example.

The General Solution

There is also a general solution (useful when the above method fails), which uses the quadratic formula:

Use that formula to get the two answers x+ and x− (one is for the "+" case, and the other is for the "−" case in the "±"), and we get this factoring:

a(x − x+)(x − x−)

Let us use the previous example to see how that works:

Example: what are the roots of 6x^2 + 5x − 6 ?

Substitute a=6, b=5 and c=−6 into the formula:

x = (−b ± √[b2 − 4ac]) / 2a

x = (−5 ± √[52 − 4×6×(−6)]) / 2×6

= (−5 ± √[25 + 144]) / 12

= (−5 ± √169) / 12

= (−5 ± 13) / 12

So the two roots are:

x+ = (-5 + 13) / 12 = 8/12 = 2/3,

x− = (-5 − 13) / 12 = −18/12 = −3/2

(Notice that we get the same result we did with the factoring we used before)

Now put those values into a(x − x+)(x − x−):

6(x − 2/3)(x + 3/2)

We can rearrange that a little to simplify it:

3(x − 2/3) × 2(x + 3/2) = (3x − 2)(2x + 3)

And we get the same factors as we did before!

## Solving

Quadratic Equations

An example of a **Quadratic Equation**:

The name **Quadratic** comes from "quad" meaning square, because the variable gets squared (like **x^2**).

It is also called an "Equation of Degree 2" (because of the "2" on the **x**)

Standard Form:

The **Standard Form** of a Quadratic Equation looks like this:

**a**,**b**and**c**are known values.**a**can't be 0.- "
**x**" is the**variable**or unknown (we don't know it yet).

Here are some more examples:

**2x^2 + 5x + 3 = 0** In this one **a=2**, **b=5** and **c=3**

**x^2 − 3x = 0** This one is a little more tricky:

- Where is
**a**? Well**a=1**, and we don't usually write "1x^2" **b = -3**- And where is
**c**? Well**c=0**, so is not shown.

**5x − 3 = 0** **Oops!** This one is **not **a quadratic equation: it is missing **x^2 **

(in other words **a=0**, which means it can't be quadratic)

Hidden Quadratic Equations!

So the "Standard Form" of a Quadratic Equation is

ax^2 + bx + c = 0

But sometimes a quadratic equation doesn't look like that! For example:

In disguise→In Standard Form (a, b and c)

**x^2 = 3x -1 **Move all terms to left hand side **x^2 - 3x + 1 = 0 (**a=1, b=-3, c=1)

**2(w^2 - 2w) = 5 **Expand (undo the brackets),

and move 5 to left **2w^2 - 4w - 5 = 0 (**a=2, b=-4, c=-5)

**z(z-1) = 3 **Expand, and move 3 to left **z^2 - z - 3 = 0 (**a=1, b=-1, c=-3)

**5 + 1/x - 1/x^2 = 0 **Multiply by x^2**5x^2 + x - 1 = 0 (**a=5, b=1, c=-1)

How To Solve It?

The "**solutions**" to the Quadratic Equation are where it is **equal to zero**.

There are usually 2 solutions (as shown in the graph above).

They are also called "**roots**", or sometimes "**zeros**"

There are 3 ways to find the solutions:

1. We can Factor the Quadratic (find what to multiply to make the Quadratic Equation)

2. We can Complete the Square, or

3. We can use the special **Quadratic Formula**:

Just plug in the values of a, b and c, and do the calculations.

We will look at this method in more detail now.

About the Quadratic Formula

Plus/Minus

First of all what is that plus/minus thing that looks like **± **?

The ± means there are TWO answers:

Here is why we can get two answers:

But sometimes we don't get two real answers, and the "Discriminant" shows why ...

Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x^2 + 6x + 1 = 0

**Coefficients are:** a = 5, b = 6, c = 1

**Quadratic Formula:** x = [ -b ± √(b^2-4ac) ] / 2a

**Put in a, b and c:** x = [ -6 ± √(62-4×5×1) ] / (2×5)

**Solve**: x = [ -6 ± √(36-20) ]/10

x = [ -6 ± √(16) ]/10

x = ( -6 ± 4 )/10

x = -0.2 **or** -1

**Answer:** x = -0.2 **or** x = -1

And we see them on this graph.

Check **-0.2**:5×(**-0.2**)² + 6×(**-0.2**) + 1

= 5×(0.04) + 6×(-0.2) + 1

= 0.2 -1.2 + 1 **= 0**

Check **-1**:5×(**-1**)² + 6×(**-1**) + 1

= 5×(1) + 6×(-1) + 1

= 5 - 6 + 1 **= 0**

Complex Solutions?

When the Discriminant (the value **b^2 - 4ac**) is negative we get Complex solutions ... what does that mean?

It means our answer will include Imaginary Numbers. Wow!

Example: Solve 5x^2 + 2x + 1 = 0

**Coefficients** are**:** a = 5, b = 2, c = 1

Note that The **Discriminant** is negative: b^2 - 4ac = 22 - 4×5×1 = **-16**

Use the **Quadratic Formula:** x = [ -2 ± √(-16) ] / 10

The square root of -16 is 4**i **(**i** is √-1)

So: x = ( -2 ± 4**i** )/10

**Answer:** x = -0.2 ± 0.4**i**

## Completing the Square

Completing the square

If we try to solve this quadratic equation by factoring,

*x^*2 + 6*x* + 2=0

we cannot. Therefore, we will complete the square. We will make the quadratic into the form

*a^*2 + 2*ab* + *b^*2=(*a* + *b*)2 .

This technique is valid only when the coefficient of *x^*2 is 1.

1) Transpose the constant term to the right

*x^*2 + 6*x* = −2.

2) Add a square number to both sides -- add the square of *half* the coefficient of *x*. In this case, add the square of half of 6; that is, add the square of 3, which is 9:

*x^*2 + 6*x* + 9 = −2 + 9.

The left-hand side is now the perfect square of (*x* + 3).

(*x* + 3)2 = 7.

3 is *half* of the coefficient 6.

That equation has the form

which implies

*a* = ±.

Therefore,

*x* + 3 = ±

*x* = −3 ±.

That is, the solutions to

*x^*2 + 6*x* + 2 = 0

are the conjugate pair,

−3 + , −3 − .

a) *x^*2 − 2*x* − 2 = 0 b) *x^*2 − 10*x* + 20 = 0

*x^*2 − 2*x* = 2 *x^*2 − 10*x* = −20

*x^*2 − 2*x* + 1 = 2 + 1 *x^*2 − 10*x* + 25 = −20 + 25

(*x* − 1)^2 = 3 (*x* − 5)^2 = 5

*x* − 1 = ± *x* − 5 = ±

*x* = 1 ± *x* = 5 ±

c) *x^*2 − 4*x* + 13 = 0 d) *x^*2 + 6*x* + 29 = 0

*x^*2 − 4*x* = −13 *x^*2 + 6*x* = −29

*x^*2 − 4*x* + 4 = −13 + 4 *x^*2 + 6*x* + 9 = −29+ 9

(*x* − 2^)2 = −9 (*x* + 3)^2 = −20

*x* − 2 = ±3*i* *x* + 3 = ±

*x* = 2 ± 3*i* *x* = −3 ± 2*i*

e) *x^*2 − 5*x* − 5 = 0 f) *x^*2 + 3*x* + 1 = 0

*x^*2 − 5*x* = 5 *x^*2 + 3*x* = −1

*x^*2 − 5*x* + 25/4 = 5 + 25/4 *x^*2 + 3*x* + 9/4 = −1 + 9/4

(*x* − 5/2)2 = 5 + 25/4 (*x* + 3/2)2 = − 1 + 9/4

*x* − 5/2 = ± *x* + 3/2 = ±

x = ± + 5/2 x = ± + 5/2

The quadratic formula

Here is a formula for finding the roots of any quadratic. It is proved by completing the square In other words, the quadratic formula completes the square for us.

*ax^*2 + *bx* + *c* = 0,

then

The two roots are on the right. One root has the plus sign; the other, the minus sign. If the square root term is irrational, then the two roots are a conjugate pair.

If we call those two roots *r*1 and *r*2 , then the quadratic can be factored as

(*x* − *r*1)(*x* − *r*2).

Those are the two roots. And they are rational. When the roots are rational, we could have solved the equation by factoring, which is always the simplest method.

3*x^*2 + 5*x* − 8 = (3*x* + 8)(*x* − 1)

*x* = −__8__

a) *x^*2 − 5*x* + 5

*a* = 1, *b* = −5, *c* = 5.

b) 2*x^*2 − 8*x* + 5

*a* = 2, *b* = −8, *c* = 5.

c) 5*x^*2 − 2*x* + 2

*a* = 5, *b* = −2, *c* = 2.

The radicand *b^*2 − 4*ac* is called the discriminant. If the discriminant is

a) **Positive**:The roots are real and conjugate.

b) **Negative**: The roots are complex and conjugate.

c) **Zero**:The roots are rational and equal -- i.e. a double root.

To prove the quadratic formula, we complete the square. But to do that, the coefficient of *x^*2 must be 1. Therefore, we will divide both sides of the original equation by *a*:

on multiplying both *c* and *a* by 4*a*, thus making the denominators the same,

This is the quadratic formula.