1. Solving Quadratic Equations by Factoring

2. Zero Factor Theorem Quadratic Equations Zero Factor Theorem
Can be written in the form ax2 + bx + c = 0.
a, b and c are real numbers and a  0.
This is referred to as standard form.
Zero Factor Theorem
If a and b are real numbers and ab = 0, then a = 0 or b = 0.
This theorem is very useful in solving quadratic equations.

3. Solving Quadratic Equations
Steps for Solving a Quadratic Equation by Factoring
Write the equation in standard form.
Factor the quadratic completely.
Set each factor containing a variable equal to 0.
Solve the resulting equations.
Check each solution in the original equation.

4. Solving Quadratic Equations
Example
Solve x2 – 5x = 24.
First write the quadratic equation in standard form.
x2 – 5x – 24 = 0
Now we factor the quadratic using techniques from the previous sections.
x2 – 5x – 24 = (x – 8)(x + 3) = 0
We set each factor equal to 0.
x – 8 = 0 or x + 3 = 0, which will simplify to
x = 8 or x = – 3
Continued.

5. Solving Quadratic Equations
Example Continued
Check both possible answers in the original equation.
82 – 5(8) = 64 – 40 = true
(–3)2 – 5(–3) = 9 – (–15) = true
So our solutions for x are 8 or –3.

6. Solving Quadratic Equations
Example
Solve 4x(8x + 9) = 5
First write the quadratic equation in standard form.
32x2 + 36x = 5
32x2 + 36x – 5 = 0
Now we factor the quadratic.
32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0
We set each factor equal to 0.
8x – 1 = 0 or 4x + 5 = 0
8x = 1 or 4x = – 5, which simplifies to x = or
Continued.

7. Solving Quadratic Equations
Example Continued
Check both possible answers in the original equation.
true
true
So our solutions for x are or

8. Finding x-intercepts
Recall earlier we found the x-intercept(s) of linear equations by letting y = 0 and solving for x. The same method works for x-intercepts in quadratic equations. Note: When the quadratic equation is written in standard form, the graph is a parabola opening up (when a > 0) or down (when a < 0), where a is the coefficient of the x2 term. The intercepts will be where the parabola crosses the x-axis.

9. Finding x-intercepts Example
Find the x-intercepts of the graph of f(x)= 4x x + 6.
The equation is already written in standard form, so we let f(x) = 0, then factor the quadratic in x.
0 = 4x2 + 11x + 6 = (4x + 3)(x + 2)
We set each factor equal to 0 and solve for x.
4x + 3 = 0 or x + 2 = 0
4x = –3 or x = –2
x = –¾ or x = –2
So the x-intercepts are the points (–¾, 0) and (–2, 0).

10. Quadratic Equations and Problem Solving

11. Strategy for Problem Solving
General Strategy for Problem Solving
Understand the problem
Read and reread the problem
Choose a variable to represent the unknown
Construct a drawing, whenever possible
Propose a solution and check
Translate the problem into an equation
Solve the equation
Interpret the result
Check proposed solution in problem
State your conclusion

12. Finding an Unknown Number
Example
The product of two consecutive positive integers is Find the two integers.
1.) Understand
Read and reread the problem. If we let
x = one of the unknown positive integers, then
x + 1 = the next consecutive positive integer.
Continued

13. Finding an Unknown Number
Example continued
2.) Translate •
The product of is =
132
two consecutive positive integers x
(x + 1)
Continued

14. Finding an Unknown Number
Example continued
3.) Solve
x(x + 1) = 132
x2 + x = (Distributive property)
x2 + x – 132 = (Write quadratic in standard form)
(x + 12)(x – 11) = (Factor quadratic polynomial)
x + 12 = 0 or x – 11 = (Set factors equal to 0)
x = –12 or x = (Solve each factor for x)
Continued

15. Finding an Unknown Number
Example continued
4.) Interpret
Check: Remember that x is suppose to represent a positive integer. So, although x = -12 satisfies our equation, it cannot be a solution for the problem we were presented.
If we let x = 11, then x + 1 = 12. The product of the two numbers is 11 · 12 = 132, our desired result.
State: The two positive integers are 11 and 12.

16. The Pythagorean Theorem
In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.
(leg a)2 + (leg b)2 = (hypotenuse)2

17. The Pythagorean Theorem
Example
Find the length of the shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg.
1.) Understand
Read and reread the problem. If we let
x = the length of the shorter leg, then
x + 10 = the length of the longer leg and
2x – 10 = the length of the hypotenuse.
Continued

18. The Pythagorean Theorem
Example continued
2.) Translate
By the Pythagorean Theorem,
(leg a)2 + (leg b)2 = (hypotenuse)2
x2 + (x + 10)2 = (2x – 10)2
3.) Solve
x2 + (x + 10)2 = (2x – 10)2
x2 + x2 + 20x = 4x2 – 40x + 100
(multiply the binomials)
2x2 + 20x = 4x2 – 40x + 100
(simplify left side)
0 = 2x2 – 60x
(subtract 2x2 + 20x from both sides)
0 = 2x(x – 30)
(factor right side)
x = 0 or x = 30
(set each factor = 0 and solve)
Continued

19. The Pythagorean Theorem
Example continued
4.) Interpret
Check: Remember that x is suppose to represent the length of the shorter side. So, although x = 0 satisfies our equation, it cannot be a solution for the problem we were presented.
If we let x = 30, then x + 10 = 40 and 2x – 10 = 50. Since = = 2500 = 502, the Pythagorean Theorem checks out.
State: The length of the shorter leg is 30 miles. (Remember that is all we were asked for in this problem.)