Quadratic Equations Solving Quadratic Equations
Mar 31, 2015
Quadratic Equations
Solving
Quadratic Equations
Quadratic Equations 27/9/2013
Solving Quadratic Equations
Standard Form
ax2 + bx + c = 0
with a ≠ 0
Solve to Find: Solutions and Solution Set
What is a solution ? What is a solution set ?
How many solutions ?
Quadratic Equations 37/9/2013
Solving Quadratic Equations Standard Form
ax2 + bx + c = 0
with a ≠ 0 Solve by:
Square Root Property Completing the Square Quadratic Formula Factoring
Quadratic Equations 47/9/2013
Solving Quadratic Equations Solutions
A solution for the equation
ax2 + bx + c = 0
is a value of variable x satisfying the equation
NOTE: For now we assume all values are real numbers
The solution set for the equation is the set of all of its solutions
Quadratic Equations 57/9/2013
Solving Quadratic Equations Examples
1. x2 + 2x – 3 = 0
2. 16x2 – 8x + 1 = 0
3. x2 + 2x + 5 = 0
Solutions: –3, 1 Solution set: {–3 , 1}
Solution: 14
Solutions: noneSolution set: { } OR
Solution set: { }14
Quadratic Equations 67/9/2013
The Square Root Property Reduced Standard Form - Type 1
ax2 = c , for a ≠ 0
Rewrite:
Square Root Property
If x2 = k then
OR
cax2 = = k
x2 = x│ │
k =x
k =x –
Quadratic Equations 77/9/2013
The Square Root Property Square Root Property
If x2 = k then
ORx2 = x│ │
k =x
k =x –Note:
This does NOT say that This DOES say that IF x2 = 4 then
OR
The Principal Root The Negative Root
= 2x = 4
2 4 =
4 –x = = –2
Quadratic Equations 87/9/2013
Square Root Property Roof repairs on a 100-story building
A tar bucket slips and falls to the sidewalk
If each story is 12.96 feet, how long does it take the bucket to hit the sidewalk?
Distance of fall: (12.96 ft/story)(100 stories) = 1296 ft
For a fall of s ft in t seconds we have
s = 16t2 OR t2 = s/16
Quadratic Equations 97/9/2013
Square Root Property Roof repairs on a 100-story building
Solving for t
We choose the positive value of t Thus
s = 16t2 OR t2 = s/16
OR =t s 4
–=t4 s
WHY ?
436
= 9= seconds=t4 s
=4
1296
Quadratic Equations 107/9/2013
Special Forms Perfect-Square Trinomials
Always the square of a binomial
A2 + 2AB + B2 = (A + B)2
A2 – 2AB + B2 = (A – B)2 … OR (B – A)2
Quadratic Equations 117/9/2013
Special Forms Examples
x2 + 6x + 9
4y2 – 20y + 25
9y2 + 12xy + 4x2
x2y2 – 6xyz + 9z2
= (x + 3)2
= (2y – 5)2
= (3y + 2x)2
= (xy – 3z)2
Quadratic Equations 127/9/2013
Completing the Square Convert to perfect-square trinomial
Solve using square root property
Example 1:
2x2 – 3x + 5 = x2 – 9x + 12
x2 + 6x = 7
To make the left side
we need to find a such that
(x + a)2 = x2 + 2ax + a2
Subtract x2 and 5, on each side, and add 9x
= x2 + 6x + a2
a perfect square
Quadratic Equations 137/9/2013
Completing the Square Convert to perfect-square trinomial
Example 1:
2x2 – 3x + 5 = x2 – 9x + 12
x2 + 6x = 7
(x + a)2 = x2 + 2ax + a2 = x2 + 6x + a2
So 2ax = 6x giving a = 3 and a2 = 9Adding 9 to each side of x2 + 6x = 7x2 + 6x + 9 = 7 + 9 = 16
(x + 3)2 = 16 x + 3 16 = + = +4
x = –3 + 4 Solution set: { –7, 1 }
Quadratic Equations 147/9/2013
Completing the Square Convert to perfect-square trinomial
Example 2:
5y2 – 87y + 89 = 25 + 15y – 4y2
9y2 – 102y = –64
For the left need to find a such that
(3y + a)2 = 9y2 + 6ay + a2
= 9y2 – 102y + a2
Add 4y2 to each side, subtract 89 and 15y
to be a perfect square we
Quadratic Equations 157/9/2013
Completing the Square Example 2:
5y2 – 87y + 89 = 25 + 15y – 4y2
9y2 – 102y = –64
(3y + a)2 = 9y2 + 6ay + a2 = 9y2 – 102y + a2
So 6ay = –102y giving a = –17 and a2 = 289
Adding 289 to each side
9y2 – 102y + 289 = –64 + 289Thus
(3y – 17)2 = 225
Quadratic Equations 167/9/2013
Completing the Square Example 2:
5y2 – 87y + 89 = 25 + 15y – 4y2
9y2 – 102y = –64
(3y – 17)2 = 225
3y – 17 225 = +15 = +
3y
= 17 15 +
y
= +17 3
15 3 =
32 3
2 3
Solution Set: { 32 3
2 3
, }
Quadratic Equations 177/9/2013
The Quadratic Formula I Complete the Square on the Standard Form
ax2 + bx + c = 0 , for a ≠ 0 Rewrite: move c, divide by a
ca
ba –x2 x + =
Now complete the square on the left side Want a constant k such that
bax2 x + + k2
is a perfect square – that is bax2 x + + k2 = x + k( )2
Quadratic Equations 187/9/2013
The Quadratic Formula II Complete the Square on the Standard Form
ca
ba –x2 x + =
bax2 x + + k2 = x + k( )2
= x2 + 2xk + k2
Thus 2k = ba … and = b
2a( )2k2
bax2 x + c
a–= + b2a( )2
+ b2a( )2
x
+ b2a( )2 c
a–= + b4a
2
2
Quadratic Equations 197/9/2013
The Quadratic Formula III Complete the Square on the Standard Form
x +( b
2a)2
= ca– + b2
4a2
+ b2
4a2= – 4ac4a2 =
4a2b2 – 4ac
x + b2a =
4a2b2 – 4ac+
x =b
2a 2ab2 – 4ac+
x = 2ab b2 – 4ac+
Square Root Property
Quadratic Formula
Quadratic Equations 207/9/2013
Quadratic Formula Examples Example 1
Solve: 7x2 + 9x + 29 = 27
First put into standard form:
7x2 + 9x + 2 = 0
x =2ab2 – 4ac ±–b Apply Quadratic
Formula
Thus a = 7 , b = 9 , c = 2
WHY ?
x = ±–9 92 – 4 7 2 ( ( ))
2 7( )
Quadratic Equations 217/9/2013
Quadratic Formula Examples Example 1
Solve: 7x2 + 9x + 29 = 27
x = ±–9 92 – 4 7 2 ( ( ))
2 7( )
=±–9 81 – 56
14 =
±–9 25 14
=±–9 5
14
Question: Are there always two solutions ?
Solution set:
{–1, } 72 –
=
–14 14
(continued)
– 4 14
Quadratic Equations 227/9/2013
Quadratic Formula Examples Example 2
Solve: 4x2 – 12x + 29 = 20
First put into standard form:
4x2 – 12x + 9 = 0 WHY ?
Thus a = 4 , b = –12 , c = 9
x =2ab2 – 4ac ±–b Apply Quadratic
Formula
x =2 4( )
–12 2 – 4 4 9 ±12 ( ( ))( )
Quadratic Equations 237/9/2013
Quadratic Formula Examples Example 2
Solve: 4x2 – 12x + 29 = 20
Solution set:
x =2 4( )
–12 2 – 4 4 9 ±12 ( ( ))( )
(continued)
x =8
144 – 144±12
= 812 x
= 23 x {
23 }
Question: Are there always two solutions ?
Quadratic Equations 247/9/2013
The Quadratic Formula The Discriminant
For ax2 + bx + c = 0 we found that
The expression b2 – 4ac is called the discriminant
It determines the number and type of solutions
x =2ab2 – 4ac ±–b
Quadratic Equations 257/9/2013
The Quadratic Formula The Discriminant
Determines the number and type of solutions
If b2 – 4ac = 0 have one real solution If b2 – 4ac > 0 have two real solutions If b2 – 4ac < 0 have two complex
solutions
Quadratic Equations 267/9/2013
Factoring Reduced Standard Form – Type 2
ax2 + bx = 0 , for a ≠ 0 Factor x and use zero-product property
x(ax + b) = 0 Zero-product Property:
if and only if p = 0 or q = 0 ... or both
For any real numbers p and q , pq = 0
Quadratic Equations 277/9/2013
Factoring Reduced Standard Form – Type 2
ax2 + bx = 0 , for a ≠ 0 Example:
Solve 3x2 – 5x = 0
x(3x – 5) = 0
From the zero-product property
x = 0 or 3x – 5 = 0 or both
Solution Set: {0 , 53 }
Quadratic Equations 287/9/2013
Special Forms Conjugate Binomials
Pairs of binomials of form (A + B) and (A – B) are called conjugate
binomials
Product of conjugate pair is always a difference of squares
(A + B)(A – B) = A2 – B2
Quadratic Equations 297/9/2013
Special Forms Conjugate Binomials
Examples
x2 – 4
9 – 4x2
4x2 – 9y2
= (x + 2)(x – 2)
= (3 – 2x)(3 + 2x)
= (2x – 3y)(2x + 3y)
Quadratic Equations 307/9/2013
Not-So-Special Forms What if the expression is not a
special form? We look for factors with integer
coefficients
In general these have form:
x2 + (a + b)x + ab = (x + a)(x + b)
We look for factors (x + a) and (x + b) that fit this form
Quadratic Equations 317/9/2013
Not-So-Special Trinomials What if the expression is not a
special form?
Factoring Examples:
x2 + 7x + 12 (x + 3)(x + 4)=
y2 + 13y + 40 (y + 5)(y + 8)=
(2x + 3)(2x + 5)=4x2 + 16x + 15
Question: Is there a systematic way to do this ?
Quadratic Equations 327/9/2013
Not-So-Special Trinomials Factoring Trinomials with negative
constant term In general these have form:
x2 + (a – b)x – ab = (x + a)(x – b)
We look for factors that fit this form
Quadratic Equations 337/9/2013
Not-So-Special Trinomials Examples
x2 – x – 12 =
y2 + 3y – 40 =
4x2 – 4x – 15 =
Note the sign of last term: always negative
Note the sign of the middle term: positive or negative
(x + 3)(x – 4)
(y – 5)(y + 8)
(2x + 3)(2x – 5)
WHY ?
WHY ?
Quadratic Equations 347/9/2013
Factor x2 – 5x – 24
x2 + (a + b)x + ab = (x + a)(x + b)
Trinomials with two negative terms
Not-So-Special Trinomials
Terms a , b of opposite sign
Sign of the larger term
Find a and b such that
Possible factors of –24 :
1, –24 2, –12 3, –8 4, –6
Since 3 – 8 = –5 , then a = 3 and b = –8
Quadratic Equations 357/9/2013
Factor x2 – 5x – 24
Trinomials with two negative terms
Not-So-Special Trinomials
Since 3 – 8 = –5 , then a = 3 and b = –8
This gives us:
NOTE: There is no solution to talk about here … since we are changing the form of an expression, NOT solving an equation
x2 – 5x – 24 = (x + 3)(x – 8)
Quadratic Equations 367/9/2013
Factor x2 + 5x – 24
x2 + (a + b)x + ab = (x + a)(x + b)
Trinomials with one negative term– the constant
Not-So-Special Trinomials
Terms a , b of opposite sign
Sign of the larger term
Find a and b such that
Possible factors of –24 :
–1, 24 –2, 12 –3, 8 –4, 6
Since 3 – 8 = 5 , then a = 8 and b = –3
Quadratic Equations 377/9/2013
Factor x2 + 5x – 24
Trinomials with one negative term– the constant
Not-So-Special Trinomials
Since 3 – 8 = 5 , then a = 8 and b = –3
This gives us:x2 + 5x – 24 = (x + 8)(x – 3)
NOTE: This is not a solution , since we arechanging the form of an expression , NOT solving an equation
Quadratic Equations 387/9/2013
Solve x2 – 8x + 12 = 0 by factoring
x2 + (a + b)x + ab = (x + a)(x + b)
Trinomials with one negative term– the middle term
Not-So-Special Trinomials
Terms a , b of same sign
Sign of both terms
Find a and b such that
Possible factors of 12 :
–1, –12 –2, –6 –3, –4
Since –2 – 6 = –8 , then a = –2 and b = –6
Quadratic Equations 397/9/2013
Solve x2 – 8x + 12 = 0 by factoring
Trinomials with one negative term– the middle term
Not-So-Special Trinomials
Since –2 – 6 = –8 , then a = –2 and b = –6
x2 – 8x + 12 = (x – 2)(x – 6) = 0 This gives us:
NOTE: This is a solution , since we aresolving an equation
So x – 2 = 0 or x – 6 = 0
Solution Set: 2, 6{ }
Quadratic Equations 407/9/2013
Solve x2 + 11x + 28 = 0 by factoring
x2 + (a + b)x + ab = (x + a)(x + b)
Trinomials with no negative terms
Not-So-Special Trinomials
Terms a , b of same sign
Sign of both terms
Find a and b such that
Possible factors of 28 :
1, 28 2, 14 4, 7
Since 4 + 7 = 11 , then a = 4 and b = 7
Quadratic Equations 417/9/2013
Solve x2 + 11x + 28 = 0 by factoring
Trinomials with no negative terms
Not-So-Special Trinomials
Since 4 + 7 = 11 , then a = 4 and b = 7
x2 + 11x + 28 = (x + 4)(x + 7) = 0This gives us:
So x + 4 = 0 or x + 7 = 0
Solution Set: – 4 , –7{ }
NOTE: This is a solution , since we aresolving an equation
Quadratic Equations 427/9/2013
Expanding a Room
A 9 by 12 room is to be expanded by the same amount x in length and width to double the area – what is x ?
9 + x
12 ft
9 ft A1 = 108
A4 = x2A3 = 12x
A2 = 9x
x ft
x ft
12 + xOriginal area:A1 = 208 ft2
New area:A = 2A2 = 216 ft2
Two methods:
A = A1 + A2 + A3 + A4
A = (9 + x)(12 + x)= 216
= 216
Quadratic Equations 437/9/2013
Expanding a Room
Two methods:A = A1 + A2 + A3 + A4
A = (9 + x)(12 + x) = 216= 216OR
Either way: x2 + 21x + 108 = 216So x2 + 21x – 108 = 0
≈ 4.3 sq. ft.
x =2 1( )
( ( ))( )± 21 2 – 4 1 –108
21 –
x =2
± 876 21 –
Quadratic Equations 447/9/2013
Think about it !