Quadratic Equation Theory

7/25/2019 Quadratic Equation Theory

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Quadratic Equation:-In elementary algebra, a quadratic equation(from the Latin

quadratus for "square") is any equationhaving the form. Where x represents an unknown, and

a, b, and c represent known numbers such that a is not equal to 0. If a = 0, then the equationis

linear, not quadratic.

Practical life example: Balls, Arrows, Missiles and Stones

If you throw a ball (or shoot an arrow, fire a missile or throw a stone) it will go

up into the air, slowing down as it goes, then come down again ...

... and a Quadratic Equation tells you where it will be!

1)An example of a Quadratic Equation:

Quadratic Equations make nice curves, like this one:
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Standard Form

The Standard Formof a Quadratic Equation looks like this:

a, band care known values. acan't be 0.

"x" is thevariableor unknown (we don't know it yet).

Here are some more examples:

2x2+ 5x + 3 =

0In this one a=2, b=5and c=3

x23x = 0 This one is a little more tricky:

Where is a? Well a=1, and we don't usually write

"1x2"

b = -3

And where is c? Well c=0, so is not shown.
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5x 3 = 0 Oops!This one is not a quadratic equation: it is

missing x2

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

Hidden Quadratic Equations!

So the "Standard Form" of a Quadratic Equation is

ax2+ bx + c = 0

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

In disguise In Standard Form a, b and c

x2= 3x 1 Move all terms to left hand side x2 3x + 1 = 0 a=1, b=3, c=1

2(w2 2w) = 5Expand(undo thebrackets),

and move 5 to left2w2 4w 5 = 0 a=2, b=4, c=5

z(z1) = 3 Expand, and move 3 to left z2 z 3 = 0 a=1, b=1, c=3

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 below above).

They are also called "roots", or sometimes "zeros"
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There are 3 ways to find the solutions:


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1. We canFactor the Quadratic(find what to multiply to make the

Quadratic Equation)

2. We canComplete 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 ...
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Discriminant

Do you see b2 4acin the formula above? It is called the Discriminant,

because it can "discriminate" between the possible types of answer:

when b2- 4acis positive, we get twoRealsolutions

when it is zero we get just ONE real solution (both answers are the

same)

when it is negative we get twoComplexsolutions

Complex solutions?Let's talk about them after we see how to use the

formula.

Using the Quadratic Formula

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

calculations.

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

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

Quadratic Formula: x = b (b2 4ac)2a

Put in a, b and c: x = 6 (62 451)25

Solve: x = 6 (3620)10

x = 6 (16)10

x = 6 410
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x = 0.2or1

Answer:x = 0.2orx = 1

And we see them on this graph.

Check -0.2: 5(0.2) + 6(0.2) + 1

= 5(0.04) + 6(0.2) + 1

= 0.2 1.2 + 1

= 0

Check -1: 5(1) + 6(1) + 1

= 5(1) + 6(1) + 1

= 5 6 + 1

= 0

Remembering the Formula

Complex Solutions?

When the Discriminant (the value b2 4ac) is negative we

get Complex solutions ... what does that mean?

It means our answer will include Imaginary Numbers . Wow!
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Example: Solve 5x + 2x + 1 = 0

Coefficientsare: a = 5, b = 2, c = 1

Note that the Discriminantis negative: b24ac = 22451 = -16

Use the Quadratic Formula: x = 2 (16)10

The square root of -16 is 4i

(iis -1, readImaginary Numbersto find out more)

So: x = 2 4i10

Answer:x = 0.2 0.4i

The graph does not cross the x-axis. That is why

we ended up with complex numbers.

In some ways it is easier: we don't need more calculation, just leave it as 0.2 0.4i.

Summary

Quadratic Equation in Standard Form: ax2+ bx + c = 0

Quadratic Equations can befactored

Quadratic Formula: x = b (b2 4ac)2a

When the Discriminant (b24ac) is:
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positive, there are 2 real solutions

zero, there is one real solution

negative, there are 2 complex solutions

Factoring Quadratics

AQuadratic Equationin Standard Form

(a, b, and ccan have any value, except that acan'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 x2+ 3x 4are:

(x+4)and (x1)

Why?Well, let us multiply them to see:

(x+4)(x1) = x(x1) + 4(x1)

= x2x + 4x 4

= x2+ 3x 4
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Multiplying (x+4)(x1)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 knowthe factors yet:

Common Factor

First check if there any common factors.

Example: what are the factors of 6x22x = 0?

6and 2have a common factor of 2:

2(3x2 x) = 0

And x2and xhave a common factor of x:

2x(3x 1) = 0
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And we have done it! The factors are 2xand 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 2x2+ 7x + 3?

No common factors.

Let us try to guessan answer, and then check if we are right ... we might get

lucky!


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

(2x+3)(x+1) = 2x2+ 2x + 3x + 3

= 2x2+ 5x + 3 (WRONG)

How about (2x+7)(x1):

(2x+7)(x1) = 2x2 2x + 7x 7= 2x2+ 5x 7(WRONG AGAIN)

OK, how about (2x+9)(x1):

(2x+9)(x1) = 2x2 2x + 9x 9= 2x2+ 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: 2x2+ 7x + 3

ac is 23 = 6and b is 7


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

In fact 6and 1do that (61=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 61=6.

Step 2: Rewrite the middle with those numbers:

Rewrite 7x with 6x and 1x:

2x2+ 6x + x+ 3

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

The first two terms 2x2+ 6xfactor into 2x(x+3)

The last two terms x+3don'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)
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Check: (2x+1)(x+3) = 2x2+ 6x + x + 3 = 2x2+ 7x + 3(Yes)

Much better than guessing!

Let us try another example:

Example: 6x2+ 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:

49 = 36 and 4+9= 5

Step 2: Rewrite 5xwith 4x and 9x:

6x2 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) = 6x2 4x + 9x 6 =6x2+ 5x 6(Yes)


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Finding Those Numbers

The hardest part is finding two numbers that multiply to give ac, and add togive 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 120and 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 rootsof 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) of6x2+ 5x 6?

We already know (from above) the factors are


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(2x + 3)(3x 2)

And we can figure out that

(2x + 3) is zero when x = 3/2

and

(3x 2) is zero whenx = 2/3

So the roots of 6x

2

+ 5x 6are:

3/2and 2/3

Here is a plot of 6x2+ 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


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

can give us clues.

Example: (continued)

Starting with 6x2+ 5x 6andjust this plot:

The roots are aroundx = 1.5 and x = +0.67, so we canguessthe roots are:

3/2and 2/3

Which can help us work out the factors 2x + 3and 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:
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a(x x+)(x x)

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

Example: what are the roots of6x2+ 5x 6?

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

x = (b [b2 4ac]) / 2a

x = (5 [52 46(6)]) / 26

= (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)


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And we get the same factors as we did before!

Completing the Square

"Completing the Square" is where we ...

... take aQuadratic Equationlike this: and turn it into this:

ax2+ bx + c = 0 a(x+d)2+ e= 0

For those of you in a hurry, I can tell you that:

and:

But if you have time, let me show you how to "Complete the Square"

yourself.

Completing the Square

Say we have a simple expression like x2+ bx. Having xtwice in the same

expression can make life hard. What can we do?

Well, with a little inspiration from Geometry we can convert it, like this:
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As you can see x2+ bxcan be rearranged nearlyinto a square ...

... and we can complete the squarewith (b/2)2

In Algebra it looks like this:

x2+ bx + (b/2)2 = (x+b/2)2

"Complete the Square"

So, by adding (b/2)2we can complete the square.

And (x+b/2)2has xonly once, which is easier to use.

Keeping the Balance

Now ... we can't just add(b/2)2without alsosubtractingit too! Otherwise

the whole value changes.

So let's see how to do it properly with an example:

Start with:

("b" is 6 in this case)


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Complete the Square:

Also subtractthe new term

Simplify it and we are done.

The result:

x2+ 6x + 7 = (x+3)22

And now xonly appears once, and our job is done!

A Shortcut Approach

Let us look at the result we want: (x+d)2+ e

When we expand (x+d)2we get x2+ 2dx + d2, so:

Now we can "force" an answer:
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We know that 6xmust end up as 2dx, so dmust be 3

Next we see that 7must become d2+ e = 9 + e, so emust be 2

And we get the same result (x+3)2 2as above!

Now, let us look at a useful application: solving Quadratic Equations ...

Solving General Quadratic Equations by Completing theSquare

We can complete the square to solvea Quadratic Equation (find where it is

equal to zero).

But a general Quadratic Equation can have a coefficient of ain front of x2:

ax2+ bx + c = 0

But that is easy to deal with ... just divide the whole equation by "a" first, then

carry on:

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

Steps

Now we can solvea Quadratic Equation in 5 steps:

Step 1Divide all terms by a(the coefficient of x2).

Step 2Move the number term (c/a) to the right side of the equation.
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Step 3Complete the square on the left side of the equation and

balance this by adding the same value to the right side of the

equation.

We now have something that looks like (x + p)

2

= q, which can be solved rathereasily:

Step 4Take the square root on both sides of the equation.

Step 5Subtract the number that remains on the left side of the

equation to findx.

Examples

Here are two examples:

Example 1: Solve x2+ 4x + 1 = 0

Step 1can be skipped in this example since the coefficient of x2is 1

Step 2Move the number term to the right side of the equation:

x2+ 4x = -1

Step 3Complete the square on the left side of the equation and balance this by

adding the same number to the right side of the equation.

(b/2)2= (4/2)2= 22= 4

x2+ 4x + 4 = -1 + 4

(x + 2)2= 3

Step 4Take the square root on both sides of the equation:

x + 2 = 3 = 1.73 (to 2 decimals)


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Step 5Subtract 2 from both sides:

x = 1.73 2 = -3.73 or -0.27

And here is an interesting and useful thing.

At the end of step 3 we had the equation:

(x + 2)2= 3

It gives us the vertex(turning point) of x2+ 4x + 1: (-

2, -3)

Example 2: Solve 5x24x 2 = 0

Step 1Divide all terms by 5

x20.8x 0.4 = 0

Step 2Move the number term to the right side of the equation:

x20.8x = 0.4

Step 3Complete the square on the left side of the equation and balance this by

adding the same number to the right side of the equation:

(b/2)2= (0.8/2)2= 0.42= 0.16

x20.8x + 0.16 = 0.4 + 0.16

(x 0.4)2= 0.56

Step 4Take the square root on both sides of the equation:

x 0.4 = 0.56 = 0.748 (to 3 decimals)


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Step 5Subtract (-0.4) from both sides (in other words, add 0.4):

x = 0.748 + 0.4 = -0.348 or 1.148

Why "Complete the Square"?

Why complete the square when we can just use the Quadratic Formula to solve

a Quadratic Equation?

Well, one reason is given above, where the new form not only shows us the

vertex, but makes it easier to solve.

There are also times when the form ax2+ bx + cmay be part of

a largerquestion and rearranging it as a(x+d)2+ emakes the solution easier,

because xonly appears once.

For example "x" may itself be a function (like cos(z)) and rearranging it may

open up a path to a better solution.

Also Completing the Square is the first step in the Derivation of the Quadratic

Formula

Just think of it as another tool in your mathematics toolbox.

Question 1Question 2Question 3Question 4Question 5Question 6Qu

estion 7Question 8Question 9Question 10

Footnote: Values of "d" and "e"

How did I get the values of dand efrom the top of the page?
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26/28

Start with

Divide the equation by a

Put c/aon other side

Add (b/2a)2to both sides


Now bring everything back...

... to the left side

... to the original multiple aof x2

And you will notice that we have got: a(x+ d)2+ e= 0

Where: , and:

Derivation of Quadratic Formula

A Quadratic Equation looks like this:

And it can be solved using the Quadratic Formula:
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27/28

That formula looks like magic, but you can follow the steps to see how it comes

about.

1. Complete the Square

ax2+ bx + c has "x" in it twice, which is hard to solve.

But there is a way to rearrange it so that "x" only appears once. It is

called Completing the Square (please read that first!).

Our aim is to get something likex2+ 2dx + d2, which can then be simplified

to (x+d)2

So, let's go:

Start with

Divide the equation by a

Put c/a on other side

Add (b/2a)2to both sides

The left hand sideis now in thex2+ 2dx + d2format, where "d" is "b/2a"

So we can re-write it this way:
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28/28


Now x only appears once and we are making progress.

2. Now Solve For "x"

Now we just need to rearrange the equation to leave "x" on the left

Start with

Square root

Move b/2a to right

That is actually solved! But let's simplify it a bit:

Multiply right by 2a/2a

Simplify:

Which is the Quadratic formula we all know and love:

Documents

Quadratic Equation Theory