1

A quadratic equation is an equation equivalent to one of the form

Where a, b, and c are real numbers and a 0 02 cbxax

To solve a quadratic equation we get it in the form above and see if it will factor.

652 xx Get form above by subtracting 5x and adding 6 to both sides to get zero on right side.-5x + 6 -5x + 6

0652 xx Factor of 6 {1,6 or 2,3}

023 xx Use the Null Factor law and set each factor = 0 and solve.

02or 03 xx 3x 2x

So if we have an equation in x and the highest power is 2, it is quadratic.

2

In this form we could have the case where b = 0.

02 cbxaxRemember standard form for a quadratic equation is:

02 cax002 cxax

When this is the case, we get the x2 alone and then square root both sides.

062 2 x Get x2 alone by adding 6 to both sides and then dividing both sides by 2

+ 6 + 6

62 2 x2 2

32 x

Now take the square root of both sides remembering that you must consider both the positive and negative root.

3x Let's check: 0632

2 0632

2

066 066

Now take the square root of both sides remembering that you must consider both the positive and negative root.

3

02 cbxaxWhat if in standard form, c = 0?

002 bxaxWe could factor by pulling an x out of each term.

032 2 xx Factor out the common x

032 xx Use the Null Factor law and set each

factor = 0 and solve.

032or 0 xx

2

3or 0 xx If you put either of these values in for

x in the original equation you can see it makes a true statement.

4

2 4 5x x

1 0x 5 0x

1 5 0x x

1x 5x

4 5x x

2 4 5 0x x

Solving Quadratic Equations by Factoring

5

Example

23 7 6x x 3 0x 3 2 0x

3 3 2 0x x

3x 2

3x

3 7 6x x

23 7 6 0x x 3 2x

6 :Factors of2, 31, 6

3:Factors of1, 3

6

Example

29 24 16x x 29 24 16 0x x

3 4 0x 3 4 3 4 0x x

4

3x

3 4x

9 16and are perfect squares

7

Example

32 18 0x x 2x

2 0x

2x

3x 3 0x 3 0x

3x 0x

2 9x 0

3x 3x 0

8

Example

23 3 20 7 0x x x

3x

3 0x

7x 7 0x 3 1 0x

1

3x

3x 3 1x

3:Factors of 1, 3 7 :Factors of 1, 7

7x 0 3 1x

9

Example

An alternate method of solving a quadratic equation is using the Principle of Taking the Square Root of Each Side of an Equation If x2 = a, then

x = + a

Solving Quadratic Equations by square roots

10

Ex: Solve by taking square roots 3x2 – 36 = 0

First, isolate x2: 3x2 – 36 = 03x2 = 36

x2 = 12

Now take the square root of both sides:

12x 2 12x

32x x 2 2 3

11

Ex: Solve by taking square roots 4(z – 3)2 = 100

First, isolate the squared factor:

4(z – 3)2 = 100(z – 3)2 = 25

Now take the square root of both sides:

25)3z( 2

253z z – 3 = + 5

z = 3 + 5

z = 3 + 5 = 8 and z = 3 – 5 = – 2 12

Ex: Solve by taking square roots 5(x + 5)2 – 75 = 0

First, isolate the squared factor:

5(x + 5)2 = 75(x + 5)2 = 15

Now take the square root of both sides:

2( x 5 ) 15 x 5 15

x 5 15 x 5 15 , x 5 15

13

Write the equation in the form Add to each side of the equation Factor the perfect-square trinomial

Take the square root of both sides of the equationSolve for x

Steps to solving Quadratic Equations by Completing the Square

14

Ex: Solve x2 + 6x + 4 = 0 by completing the square

First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1.

x2 + 6x = – 4Add to both sides:

b =

6

6 = 32 = 9

x2 + 6x + 9 = – 4 + 9

x2 + 6x + 9 = 5

(x + 3)2 = 5

Now take the square root of both sides

15

2( 3) 5x

3 5x

3 5x

{ 3 5, 3 5}x

16

solve by completing the square.

02162 2 xx To complete the square we want the coefficient of the x2 term to be 1.

Divide everything by 20182 xx2 2 2 2

Since it doesn't factor get the constant on the other side ready to complete the square.

___1___82 xx

So what do we add to both sides?16

16 16

Factor the left hand side 24 15x

Square root both sides (remember ) 154 2 x

154 x 154xAdd 4 to both sides to get x alone

2

2

8

the middle term's coefficient divided by 2 and squared

17

Ex: Solve 2y2 = 3 – 5y by completing the square

First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1.

2y2 + 5y = 3

Add [½(b)]2 to both sides: b =

(5/2)

5/2 [½(5/2)]2 = (5/4)2 = 25/16

y2 + (5/2)y + 25/16 = (3/2) + 25/16

y2 + (5/2)y + 25/16 = 24/16 + 25/16

(y + 5/4)2 = 49/16

Now take the square root of both sides

y2 + (5/2)y = (3/2)

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2( 5 / 4) 49 /16y

5 / 4 (7 / 4)y

(5 / 4) (7 / 4)y

y= - (5/4) + (7/4) = 2/4 = ½

and y= - (5/4) - (7/4) = -12/4 = - 3

y= { ½ , - 3}

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Ex: Solve 3n– 5 = (n – 1)(n – 2)

Is this a quadratic equation?

3n – 5 = n2 – 2n – n + 2

3n – 5 = n2 – 3n + 2

n2 – 6n + 7 = 0

Collect all terms

n2 – 6n = – 7 [½(-6)]2 = (-3)2 = 9

n2 – 6n + 9 = – 7 + 9

(n – 3)2 = 2

20

2( 3) 2n

3 2n 3 2n

{3 2,3 2}n

21

Consider a quadratic equation of the form ax2 + bx + c = 0 for (a≠0).

Completing the square

2ax bx c 2

b cx x

a a

2 2

2

2 2

b b c bx x

a 4a a 4a

Solving Quadratic Equations by Quadratic Formula

22

Solutions to ax2 + bx + c = 0 for (a≠0) are:

22

2

b b 4acx

2a 4a

2b b 4acx

2a

2 2

2

2 2 2

b b 4ac bx x

a 4a 4a 4a

23

WHY USE THE QUADRATIC FORMULA?

The quadratic formula allows you to solve

any quadratic equation, even if you cannot

factor it.

An important piece of the quadratic formula

is what’s under the radical:

b2 – 4ac

This piece is called the discriminant.

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WHY IS THE DISCRIMINANT IMPORTANT?

The discriminant will take on a value that is positive,

0, or negative.

The value of the discriminant indicates two distinct

real solutions, one real solution, or no real solutions,

respectively. 25

WHAT THE DISCRIMINANT TELLS YOU!

Value of the Discriminant Nature of the Solutions

<0 Negative 2 imaginary solutions

=0 Zero 2- Double Solution

>0 Positive – perfect square 2 Real's- Rational

>0 Positive – non-perfect square 2 Real's- Irrational

26

27

Use the discriminant to determine the number and

type of solutions for the following equation.

5 – 4x + 12x2 = 0

a = 12, b = –4, and c = 5b2 – 4ac = (–4)2 – 4(12)(5)

= 16 – 240 = –224<0

There are no real solutions.

Example

28

6 – 3x2 –2x = 0

a = – 3, b = –2, and c = 6b2 – 4ac = (–2)2 – 4(– 3)(6)

= 4 + 72 = 76>0

There are 2 distinct root .

Example

29

x2 –4x +4= 0

a = 1, b = – 4, and c = 4b2 – 4ac = (–4)2 – 4(1)(4)

= 16 – 16 = 0

We have two double root

Example

By completing the square on a general quadratic equation in standard form we come up with what is called the quadratic formula.

a

acbbx

2

42

This formula can be used to solve any quadratic equation whether it factors or not. If it factors, it is generally easier to factor---but this formula would give you the solutions as well.

We solved this by completing the square but let's solve it using the quadratic formula

a

acbbx

2

42

1

(1)

(1)

6 6 (3)2

12366

Don't make a mistake with order of operations! Let's do the power and the multiplying first.

02 cbxax

0362 xx

30

2

12366 x

2

246

626424

2

626

2

632

There's a 2 in common in the terms of the numerator

63 These are the solutions we got when we completed the square on this problem.

NOTE: When using this formula if you've simplified under the radical and end up with a negative, there are no real solutions.(There are complex (imaginary) solutions, but that will be dealt with in year 12 Calculus).

31

ax2 + bx + c = 0 for (a≠0).

If,then

If , then we have two complex number (no-Real –Root)

If ,then

2

2

b b 4ac bx'

2a 2a

b b 4ac bandx"

2a 2a

bx' x"

2a

=b2 – 4ac

32

If,then

If ,then we have two complex number (no-Real –Root)

If ,then

b' b' ac b' 'x'

a a

b' b' ac b' 'andx"

a a

b'x' x"

a

=–ac where =

If b is an even number in the quadratic Equation ax2 + bx + c = 0 for

(a≠0).

33

The sum of the roots of the Quadratic equation:

ax2 + bx + c = 0 is

And the product of roots is

This means that:

Sum and Product of the Root

𝑺=−𝒃𝒂

{𝑥′+𝑥 ′ ′=𝑆=−𝑏𝑎

𝑥 ′ .𝑥 ′ ′=𝑃=𝑐𝑎

P

34

35

Note:Two numbers whose given sum is S and given product is P are roots of the equation x2 - Sx + P= 0 Example:

Find two numbers knowing their sum S and their product P :

S=4 and P=4

The two required numbers are the solutions of the equation: x2 - 4x + 4= 0 ,that is (x-2)2

=0,which gives x’=x”=2

36

Example

Write the form of a quadratic equation whose roots are x’= and x”=Solution:The sum S=x’+x”=

And the product P=x’.x”=

The equation :

Example

22 7 11 0x x

2

2

4

2

7 7 4(2)( 11

2, 7, 11

7 137 2 Reals - Irr

)

2

ational4

(2)

b b ac

b

a

a c

Solve using the Quadratic Formula

37

Ex: Use the Quadratic Formula to solve x2 + 7x + 6 = 0

Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by

a2ac4bb

x2

Identify a, b, and c in ax2 + bx + c = 0:

a = b = c = 1 7 6

Now evaluate the quadratic formula at the identified values of a, b, and c

38

)1(2)6)(1(477

x2

224497

x

2257

x

257

x

x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6

x = { - 1, - 6 }

39

Ex: Use the Quadratic Formula to solve

2m2 + m – 10 = 0

Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by

a

acbbm

2

42

Identify a, b, and c in am2 + bm + c = 0:

a = b = c = 2 1 - 10

Now evaluate the quadratic formula at the identified values of a, b, and c

40

)2(2)10)(2(411

m2

48011

m

4811

m

491

m

m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2

m = { 2, - 5/2 }

41

42

Solve 11n2 – 9n = 1 by the quadratic formula.11n2 – 9n – 1 = 0, soa = 11, b = -9, c = -1

)11(2

)1)(11(4)9(9 2

n

22

44819

22

1259

22

559

Example

43

)1(2

)20)(1(4)8(8 2

x

2

80648

2

1448

2

128 20 4 or , 10 or 22 2

x2 + 8x – 20 = 0 (multiply both sides by 8)

a = 1, b = 8, c = 20

8

1

2

5Solve x2 + x – = 0 by the quadratic formula.

Example

44

Solve x(x + 6) = 30 by the quadratic formula.x2 + 6x + 30 = 0

a = 1, b = 6, c = 30

)1(2

)30)(1(4)6(6 2

x

2

120366

2

846

So there is no real solution.

Example

45

Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1)

Let a = 1, b = -3, c = 1

)1(2

)1)(1(4)3(3 2

x

2

493

2

53

Example

46

Solve the following quadratic equation.

02

1

8

5 2 mm

0485 2 mm

0)2)(25( mm

02025 mm or

25

2 mm or

Example

Solving Quadratic Equations by the Quadratic Formula

2

2

2

2

2

1. 2 63 0

2. 8 84 0

3. 5 24 0

4. 7 13 0

5. 3 5 6 0

x x

x x

x x

x x

x x

Try the following examples. Do your work on your paper and then check your answers.

1) { 9,7}

2) {6, 14}

3) { 3,8}

7 3 7 34) ,

2 2

5 47 5 475) ,

6 6

x

x

x

i ix

i ix

47

0

A cliff diver is 64 feet above the surface of the water. The formula for calculating the height (h) of the diver after t seconds is: 216 64.h t How long does it take for the diver to hit the surface of the water?

0 0

2 0t 2 0t 2t 2t seconds

216 64t 16 2 4t

16 2t 2t

Quadratic Equations and Problem Solving

48

5 176w w

The length of a rectangular garden is 5 feet more than its width. The area of the garden is 176 square feet. What are the length and the width of the garden?

11w The width is w.

11 0w 11w

The length is w+5.l w A

2 5 176w w 2 5 176 0w w

16 0w 16w

11w 11 5l 16l

feet

feet

176 :Factors of1,176 2, 88 4, 44

8, 22 11,16

16w 0

Quadratic Equations and Problem Solving

49

x

Find two consecutive odd numbers whose product is 23 more than their sum?

Consecutive odd numbers: x

5x 5x 2 2 2 25x x x

2 25 0x 5x

5 0x 5 0x

5, 3 5, 7

5 2 3 5 2 7

2.x 2x 2x x 23

2 22 2 2 25xx x x x

2 25 2525x

2 25x

5x 0

Quadratic Equations and Problem Solving

50

a x

The length of one leg of a right triangle is 7 meters less than the length of the other leg. The length of the hypotenuse is 13 meters. What are the lengths of the legs?

12a

.Pythagorean Th

22 27 13x x

5x

5

meters

7b x 13c

2 2 14 49 169x x x 22 14 120 0x x 22 7 60 0x x

2

5 0x 12 0x

12x

12 7b meters

2 2 2a b c

60 :Factors of 1, 60 2, 303, 20 4,15 5,12

5x 12x 0

6,10

Quadratic Equations and Problem Solving

51

2x

The square of a number minus twice the number is 63. Find the number.

7x

7x

x is the number.

2 2 63 0x x

7 0x 9 0x

9x

2x 63

63:Factors of 1, 63 3, 21 7, 9

9x 0

Quadratic Equations and Problem Solving

52

SUMMARY OF SOLVING QUADRATIC EQUATIONS

• Get the equation in standard form: 02 cbxax

• If there is no middle term (b = 0) then get the x2 alone and square root both sides (if you get a negative under the square root there are no real solutions).• If there is no constant term (c = 0) then factor out the common x and use the null factor law to solve (set each factor = 0).

• If a, b and c are non-zero, see if you can factor and use the null factor law to solve.

• If it doesn't factor or is hard to factor, use the quadratic formula to solve (if you get a negative under the square root there are no real solutions).

53

a

acbbxcbxax

2

40

22

If we have a quadratic equation and are considering solutions from the real number system, using the quadratic formula, one of three things can happen.

3. The "stuff" under the square root can be negative and we'd get no real solutions.

The "stuff" under the square root is called the discriminant.

This "discriminates" or tells us what type of solutions we'll have.

1. The "stuff" under the square root can be positive and we'd get two unequal real solutions 04 if 2 acb

2. The "stuff" under the square root can be zero and we'd get one solution (called a repeated or double root because it would factor into two equal factors, each giving us the same solution).

2if 4 0b ac

04 if 2 acb

54