The document discusses solving quadratic equations using the quadratic formula. It defines the discriminant as the expression under the radical sign in the quadratic formula, and explains that the discriminant determines the number of real roots: a positive discriminant means two real roots, zero discriminant means one real root, and a negative discriminant means no real roots. Examples are provided to demonstrate solving quadratic equations using the quadratic formula and interpreting the discriminant.

1. Chapter 16
The Quadratic Formula
 Algebra 1
 Mr. Swartz

2. Proving the Quadratic Formula
 Solve by completing the square
02
 cbxax
a
c
x
a
b
x 2
2
22
42 a
b
a
b






a
c
a
b
a
b
x
a
b
x  2
2
2
2
2
44
a
c
a
b
a
b
x 





 2
22
42
2
22
4
4
2 a
acb
a
b
x








a
acb
a
b
x
2
4
2
2


22
22
4
4
42 a
ac
a
b
a
b
x 






2
4LCM a
a
acbb
x
2
42


a
acb
a
b
x
2
4
2
2



3. a2
Quadratic Formula
 There was a negative boy who couldn’t decide to
go to this radical party. Because the boy was
square, he lost out on 4 awesome chicks so he
cried his way home when it was all over at 2 AM.
b  2
b ac4
x

4. VIDEO

5. Quadratic Formula
a
acbb
x
2
42


02
 cbxax
IF
THEN

6. #1 Solve using the quadratic formula.
0273 2
 xx
a
acbb
x
2
42


2,7,3  cba
)3(2
)2)(3(4)7()7( 2

x
6
24497 
x
6
57 
x
6
12
x
6
2
x
3
1
,2x
6
257 
x

7. #2 Solve using the quadratic formula
542 2
 xx
a
acbb
x
2
42


5,4,2  cba
)2(2
)5)(2(4)4()4( 2

x
4
40164 
x
4
564 
x
4
1424 
x
0542 2
 xx
2
14
1x
9.2x 9.0x
4
1444 
x

8. Solve using the quadratic formula
3.
4.
mm 1083 2

xx 16642


9. #4 Solve using the quadratic formula
mm 1083 2

a
acbb
m
2
42


8,10,3  cba
)3(2
)8)(3(4)10()10( 2

m
6
9610010 
m
6
1410
m
6
24
m
6
4
m
3
2
,4 m
08103 2
 mm
6
19610 
m

10. #5 Solve using the quadratic formula
xx 16642

a
acbb
x
2
42


64,16,1  cba
 
)1(2
)64)(1(41616
2

x
2
25625616 
x
2
016 
x
8x
064162
 xx

11. Solve 11n2 – 9n = 1 by the quadratic
formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1



)11(2
)1)(11(4)9(9 2
n 

22
44819


22
1259
22
559 
The Quadratic Formula
Example

12. 


)1(2
)20)(1(4)8(8 2
x 

2
80648


2
1448


2
128 20 4
or , 10 or 2
2 2


x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
8
1
2
5
Solve x2 + x – = 0 by the quadratic formula.
The Quadratic Formula
Example

13. 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.
The Quadratic Formula
Example

14. 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
Solving Equations
Example
GCF is 4
By factoring out a GCF it helps
by making your a, b, and c smaller

15. Solve the following quadratic equation.
0
2
1
8
5 2
 mm
0485 2
 mm
0)2)(25(  mm
02025  mm or
2
5
2
 mm or
Solving Equations
Example
ELIMINATE FRACTIONS
Multiply by the GCF, 8
Wait, it factors, a*c
5* -4 = -20 Factors that add
To +8 are +10, -2.
*You can solve by the Quadratic
Formula if you prefer*

16. Solving Quadratic Equations
Steps in Solving Quadratic Equations
1) If the equation is in the form (ax+b)2 = c, use
the square root property to solve.
2) If not solved in step 1, write the equation in
standard form.
3) Try to solve by factoring.
4) If you haven’t solved it yet, use the quadratic
formula.

17. The Discriminant
 Discriminant  In the quadratic
formula, the expression
underneath the radical
that describes the
nature of the roots.
a
acbb
x
2
42


acb 4ntdiscrimina 2


18. Understanding the discriminant
Discriminant
acb 42

# of real roots
042
 acb 2 real roots
042
 acb 1 real roots
042
 acb No real roots

19. #6 Using the discriminant
0134 2
 yy
acb 4ntdiscrimina 2

1,3,4  cba
)1)(4(4)3(ntdiscrimina 2

169ntdiscrimina 
52ntdiscrimina 
052 
rootsreal2

20. #7 Using the discriminant
54 2
 xx
acb 4ntdiscrimina 2

5,1,4  cba
)5)(4(4)1(ntdiscrimina 2

801ntdiscrimina 
79ntdiscrimina 
079
rootsreal0

21. Using the discriminant
8.
9.
542 2
 xx
484 2
 xx
56ntdiscrimina 
rootsreal2
0ntdiscrimina 
rootreal1

22. #8 Using the discriminant
542 2
 xx
acb 4ntdiscrimina 2

0542 2
 xx
)5)(2(4)4(ntdiscrimina 2

4061ntdiscrimina 
56ntdiscrimina 
056 
rootsreal2

23. #9 Using the discriminant
484 2
 xx
acb 4ntdiscrimina 2

0484 2
 xx
)4)(4(4)8(ntdiscrimina 2

6446ntdiscrimina 
0ntdiscrimina 
00 
rootreal1