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 6x and 1x:

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 4i (i is √-1)

So: x = ( -2 ± 4i )/10


Answer: x = -0.2 ± 0.4i

Completing the Square

Completing the square


If we try to solve this quadratic equation by factoring,

x^2 + 6x + 2=0

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

a^2 + 2ab + 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 + 6x = −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 + 6x + 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

a^2 = b

which implies

a = ±.

Therefore,

x + 3 = ±

x = −3 ±.

That is, the solutions to

x^2 + 6x + 2 = 0

are the conjugate pair,

−3 + , −3 − .

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

x^2 − 2x = 2 x^2 − 10x = −20

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

(x − 1)^2 = 3 (x − 5)^2 = 5

x − 1 = ± x − 5 = ±

x = 1 ± x = 5 ±

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

x^2 − 4x = −13 x^2 + 6x = −29

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

(x − 2^)2 = −9 (x + 3)^2 = −20

x − 2 = ±3i x + 3 = ±

x = 2 ± 3i x = −3 ± 2i

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

x^2 − 5x = 5 x^2 + 3x = −1

x^2 − 5x + 25/4 = 5 + 25/4 x^2 + 3x + 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 r1 and r2 , then the quadratic can be factored as

(xr1)(xr2).


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.

3x^2 + 5x − 8 = (3x + 8)(x − 1)

x = −8

a) x^2 − 5x + 5

a = 1, b = −5, c = 5.


b) 2x^2 − 8x + 5

a = 2, b = −8, c = 5.


c) 5x^2 − 2x + 2

a = 5, b = −2, c = 2.

The radicand b^2 − 4ac 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 4a, thus making the denominators the same,

This is the quadratic formula.

Quadratic Equations - Factoring and Quadratic Formula
❤² How to Solve Quadratic Equations By Factoring (mathbff)