Lecture quadratic equations good one

QUADRATIC
EQUATIONS
MSJC ~ San Jacinto Campus
Math Center Workshop Series
Janice Levasseur

Basics
• A quadratic equation is an equation
equivalent to an equation of the type
ax2 + bx + c = 0, where a is nonzero
• We can solve a quadratic equation by
factoring and using The Principle of Zero
Products
If ab = 0, then either a = 0, b = 0, or both a
and b = 0.

Ex: Solve (4t + 1)(3t – 5) = 0
Notice the equation as given is of the form ab = 0
 set each factor equal to 0 and solve
4t + 1 = 0 Subtract 1
4t = – 1 Divide by 4
t = – ¼
3t – 5 = 0 Add 5
3t = 5 Divide by 3
t = 5/3
Solution: t = - ¼ and 5/3  t = {- ¼, 5/3}

Ex: Solve x2 + 7x + 6 = 0
Quadratic equation  factor the left hand side (LHS)
x2 + 7x + 6 = (x + 6 )(x + 1 )
 x2 + 7x + 6 = (x + 6)(x + 1) = 0
Now the equation as given is of the form ab = 0
x + 6 = 0
x = – 6
x + 1 = 0
x = – 1
Solution: x = - 6 and – 1  x = {-6, -1}

Ex: Solve x2 + 10x = – 25
Quadratic equation but not of the form ax2 + bx + c = 0
Add 25  x2 + 10x + 25 = 0
x2 + 10x + 25 = (x + 5 )(x + 5 )
 x2 + 10x + 25 = (x + 5)(x + 5) = 0
x + 5 = 0
x = – 5
x + 5 = 0
x = – 5
Solution: x = - 5  x = {- 5}  repeated root

Ex: Solve 12y2 – 5y = 2
Subtract 2  12y2 – 5y – 2 = 0
ac method  a = 12 and c = – 2
ac = (12)(-2) = - 24  factors of – 24 that sum to - 5
1&-24, 2&-12, 3&-8, . . . 
 12y2 – 5y – 2 = 12y2 + 3y – 8y – 2
= 3y(4y + 1) – 2(4y + 1)
= (3y – 2)(4y + 1)

 12y2 – 5y – 2 = 0
 12y2 – 5y – 2 = (3y - 2)(4y + 1) = 0
3y – 2 = 0
3y = 2
4y + 1 = 0
4y = – 1
y = 2/3 y = – ¼
Solution: y = 2/3 and – ¼  y = {2/3, - ¼ }

Ex: Solve 5x2 = 6x
Subtract 6x  5x2 – 6x = 0
5x2 – 6x = x( 5 x – 6 )
 5x2 – 6x = x(5x – 6) = 0
x = 0
5x – 6 = 0
5x = 6
x = 6/5
Solution: x = 0 and 6/5  x = {0, 6/5}

Solving by taking square roots
• 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

Ex: Solve by taking square roots 3x2 – 36 = 0
First, isolate x2: 3x2 – 36 = 0
3x2 = 36
x2 = 12
Now take the square root of both sides:
x 12 2 
x   12
x   2  2  3
x   2 3

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
(z 3) 25 2  
z  3   25
z – 3 = + 5
z = 3 + 5
 z = 3 + 5 = 8 and z = 3 – 5 = – 2

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
( x  5 )2  15
x  5   15
x  5  15
x  5  15 , x  5  15

Completing the Square
• Recall from factoring that a Perfect-Square
Trinomial is the square of a binomial:
Perfect square Trinomial Binomial Square
x2 + 8x + 16 (x + 4)2
x2 – 6x + 9 (x – 3)2
• The square of half of the coefficient of x
equals the constant term:
( ½ * 8 )2 = 16
[½ (-6)]2 = 9

Completing the Square
• Write the equation in the form x2 + bx = c
• Add to each side of the equation [½(b)]2
• Factor the perfect-square trinomial
x2 + bx + [½(b)] 2 = c + [½(b)]2
• Take the square root of both sides of the
equation
• Solve for x

Ex: Solve w2 + 6w + 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.
w2 + 6
6w = – 4
Add [½(b)]2 to both sides: b =
6  [½(6)]2 = 32 = 9
w2 + 6w + 9 = – 4 + 9
w2 + 6w + 9 = 5
(w + 3)2 = 5
Now take the square root of both sides

5 ) 3 w ( 2  
w  3   5
w  3 5
w  {3  5,3  5}

Ex: Solve 2r2 = 3 – 5r by completing the square
First, rewrite the equation with the constant on one
side of the equals and a lead coefficient of 1.
2r2 + 5r = 3
 r2 + (5/2)
r = (3/2)
Add [½(b)]2 to both sides: b =
5/2 [½(5/2)]2 = (5/4)2
= 25/16
r2 + (5/2)r + 25/16 = (3/2) + 25/16
r2 + (5/2)r + 25/16 = 24/16 + 25/16
(r + 5/4)2 = 49/16
Now take the square root of both sides

(r 5/ 4) 49/16 2  
r  5/ 4  (7/ 4)
r  (5/ 4)  (7/ 4)
r = - (5/4) + (7/4) = 2/4 = ½
and r = - (5/4) - (7/4) = -12/4 = - 3
r = { ½ , - 3}

Ex: Solve 3p – 5 = (p – 1)(p – 2)
Is this a quadratic equation? FOIL the RHS
3p – 5 = p2 – 2p – p + 2
3p – 5 = p2 – 3p + 2
p2 – 6p + 7 = 0
Collect all terms
A-ha . . .
Quadratic Equation  complete the square
p2 – 6p = – 7  [½(-6)]2 = (-3)2 = 9
p2 – 6p + 9 = – 7 + 9
(p – 3)2 = 2

2 ) 3 p( 2  
p  3   2
p  3  2
p  {3  2,3  2}

The Quadratic Formula
• Consider a quadratic equation of the form
ax2 + bx + c = 0 for a nonzero
• Completing the square
ax2  bx   c

b c
x 2  x

a a
b b c b
2 2
2
    
2 2
x x
a 4a a 4a

The Quadratic Formula
b b 4ac b
2 2
    
2
  
   
 
b b 2
4ac
2
x
2a 4a
Solutions to ax2 + bx + c = 0 for a nonzero
are
  

2 b b 4ac
x
2a
2
2 2 2
x x
a 4a 4a 4a

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
b b 4ac
2a
x
2   

Identify a, b, and c in ax2 + bx + c = 0:
a = 1 b = c =
1
7
7
6
Now evaluate the quadratic formula at the identified
values of a, b, and c

7 7 4(1)(6)
2(1)
x
2   

7 49 24
2
x
  

7 25
2
x
 

7 5
2
x
 

x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6
x = { - 1, - 6 }

Ex: Use the Quadratic Formula to solve
1
2m2 + m = 0
2
Recall: For quadratic equation ax2 + bx + c = 0,
the solutions to a quadratic equation are given by
4 2   

b b ac
a
m
2
Identify a, b, and c in am2 + bm + c = 0:
a = 2 b = 1
c =
- 10
– 10
Now evaluate the quadratic formula at the identified
values of a, b, and c

1 1 4(2)( 10)
2(2)
m
2    

1 1 80
4
m
  

1 81
4
m
 

1 9
4
m
 

m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2
m = { 2, - 5/2 }

Lecture quadratic equations good one

Lecture quadratic equations good one

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Lecture quadratic equations good one