1. 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.

2. Notice the equation as given is of the form ab = 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}

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

4. Quadratic equation but not of the form ax2 + bx + c = 0
Ex: Solve x2 + 10x = – 25
Quadratic equation but not of the form ax2 + bx + c = 0
 x2 + 10x + 25 = 0
Add 25
Quadratic equation  factor the left hand side (LHS)
x2 + 10x + 25 = (x )(x ) 5 5
 x2 + 10x + 25 = (x + 5)(x + 5) = 0
Now the equation as given is of the form ab = 0
 set each factor equal to 0 and solve
x + 5 = 0
x + 5 = 0
x = – 5
x = – 5
Solution: x = - 5  x = {- 5}  repeated root

5. Quadratic equation but not of the form ax2 + bx + c = 0
Ex: Solve 12y2 – 5y = 2
Quadratic equation but not of the form ax2 + bx + c = 0
 12y2 – 5y – 2 = 0
Subtract 2
Quadratic equation  factor the left hand side (LHS)
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)

6.  12y2 – 5y – 2 = 0
 12y2 – 5y – 2 = (3y - 2)(4y + 1) = 0
Now the equation as given is of the form ab = 0
 set each factor equal to 0 and solve
3y – 2 = 0
4y + 1 = 0
3y = 2
4y = – 1
y = 2/3
y = – ¼
Solution: y = 2/3 and – ¼  y = {2/3, - ¼ }

7. Quadratic equation but not of the form ax2 + bx + c = 0  5x2 – 6x = 0
Ex: Solve 5x2 = 6x
Quadratic equation but not of the form ax2 + bx + c = 0
 5x2 – 6x = 0
Subtract 6x
Quadratic equation  factor the left hand side (LHS)
5x2 – 6x = x( ) 5x
– 6
 5x2 – 6x = x(5x – 6) = 0
Now the equation as given is of the form ab = 0
 set each factor equal to 0 and solve
5x – 6 = 0
x = 0
5x = 6
x = 6/5
Solution: x = 0 and 6/5  x = {0, 6/5}

9. 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 = +

10. 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:

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:
z – 3 = + 5
z = 3 + 5
 z = = 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:

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

14. 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

15. 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 + 6w = – 4 6
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

17. 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)
(5/2)
Add [½(b)]2 to both sides:
b =
5/2
[½(5/2)]2 = (5/4) = 25/16
r2 + (5/2)r + 25/16 = (3/2) + 25/16
r2 + (5/2)r + 25/16 = 24/ /16
(r + 5/4)2 = 49/16
Now take the square root of both sides

18. r = - (5/4) + (7/4) = 2/4 = ½
and r = - (5/4) - (7/4) = -12/4 = - 3
r = { ½ , - 3}

19. 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
Collect all terms
A-ha . . .
Quadratic Equation  complete the square
p2 – 6p + 7 = 0
p2 – 6p = – 7
 [½(-6)]2 = (-3)2 = 9
p2 – 6p + 9 = – 7 + 9
(p – 3)2 = 2

21. The Quadratic Formula
Consider a quadratic equation of the form ax2 + bx + c = 0 for a nonzero
Completing the square

22. The Quadratic Formula
Solutions to ax2 + bx + c = 0 for a nonzero are

23. Ex: Use the Quadratic Formula to solve x2 + 7x + 6 = 01 7 6
Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by
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

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

25. Ex: Use the Quadratic Formula to solve
2m2 + m – 10 = 0 2 1
– 10
Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by
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

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

27. Any questions . . .