1. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics
§8.2 Quadratic Equation
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. 8.1 Review § Any QUESTIONS About Any QUESTIONS About HomeWork
MTH 55
Review §
Any QUESTIONS About
§8.1 → Complete the Square
Any QUESTIONS About HomeWork
§8.1 → HW-31

3. This is one of the MOST FAMOUS Formulas in all of Mathematics
The Quadratic Formula
The solutions of ax2 + bx + c = 0 are given by
This is one of the MOST FAMOUS Formulas in all of Mathematics

4. §8.2 Quadratic Formula The Quadratic Formula
Problem Solving with the Quadratic Formula

5. Derive Quadratic Formula - 1
Consider the General Quadratic Equation
Next, Divide by “a” to give the second degree term the coefficient of 1
Where a, b, c are CONSTANTS
Solve This Eqn for x by Completing the Square
First; isolate the Terms involving x
Now add to both Sides of the eqn a “quadratic supplement” of (b/2a)2

6. Derive Quadratic Formula - 2
Now the Left-Hand-Side (LHS) is a PERFECT Square
Use the Perfect Sq Expression
Finally Use the Sq-Root Property to solve for x
Solve for x; but first let

7. Derive Quadratic Formula - 3
Alternatively, Start with the PERFECT SQUARE Expression
Combine Terms inside the Radical over a Common Denom
Take the Square Root of Both Sides

8. Derive Quadratic Formula - 4
Note that Denom is, itself, a PERFECT SQ
Now Combine over Common Denom
But this the Renowned QUADRATIC FORMULA
Note That it was DERIVED by COMPLETING the SQUARE
Next, Isolate x
The quadratic formula was derived BEFORE Phythagorus

9. Example a) 2x2 + 9x − 5 = 0
Solve using the Quadratic Formula: 2x2 + 9x − 5 = 0
Soln a) Identify a, b, and c and substitute into the quadratic formula:
2x2 + 9x − 5 = 0
a b c
Now Know a, b, and c

10. Solution a) 2x2 + 9x − 5 = 0 Using a = 2, b = 9, c = −5
Recall the Quadratic Formula → Sub for a, b, and c
Be sure to write the fraction bar ALL the way across.

11. Solution a) 2x2 + 9x − 5 = 0
From Last Slide:
So:
The Solns:

12. Example b) x2 = −12x + 4
Soln b) write x2 = −12x + 4 in standard form, identify a, b, & c, and solve using the quadratic formula:
1x2 + 12x – 4 = 0
a b c

13. Example c) 5x2 − x + 3 = 0
Soln c) Recognize a = 5, b = −1, c = 3 → Sub into Quadratic Formula
The COMPLEX No. Soln
Since the radicand, –59, is negative, there are NO real-number solutions.

14. Quadratic Equation Graph
The graph of a quadratic eqn describes a “parabola” which has one of a:
Bowl shape
Dome shape
x intercepts
vertex
The graph, depending on the “Vertex” Location, may have different numbers of of x-intercepts: 2 (shown), 1, or NONE

15. The Discriminant
It is sometimes enough to know what type of number (Real or Complex) a solution will be, without actually solving the equation.
From the quadratic formula, b2 – 4ac, is known as the discriminant.
The discriminant determines what type of number the solutions of a quadratic equation are.
The cases are summarized on the next sld

16. Soln Type by Discriminant
Discriminant b2 – 4ac
Nature of Solutions
x-Intercepts
Only one solution; it is a real number
Only one
Positive
Two different real-number solutions
Two different
Negative
Two different NONreal complex-number solutions (complex conjugates)
None

17. Example  Discriminant
Determine the nature of the solutions of:
5x2 − 10x + 5 = 0
SOLUTION
Recognize a = 5, b = −10, c = 5
Calculate the Discriminant
b2 − 4ac = (−10)2 − 4(5)(5) = 100 − 100 = 0
There is exactly one, real solution.
This indicates that 5x2 − 10x + 5 = 0 can be solved by factoring  5(x − 1)2 = 0

18. Example  Discriminant
Determine the nature of the solutions of:
5x2 − 10x + 5 = 0
SOLUTION Examine Graph
Notice that the Graph crosses the x-axis (where y = 0) at exactly ONE point as predicted by the discriminant

19. Example  Discriminant
Determine the nature of the solutions of:
4x2 − x + 1 = 0
SOLUTION
Recognize a = 4, b = −1, c = 1
Calculate the Discriminant
b2 – 4ac = (−1)2 − 4(4)(1) =1 − 16 = −15
Since the discriminant is negative, there are two NONreal complex-number solutions

20. Example  Discriminant
Determine the nature of the solutions of:
4x2 − 1x + 1 = 0
SOLUTION Examine Graph
Notice that the Graph does NOT cross the x-axis (where y = 0) indicating that there are NO real values for x that satisfy this Quadratic Eqn

21. Example  Discriminant
Determine the nature of the solutions of:
2x2 + 5x = −1
SOLUTION: First write the eqn in Std form of ax2 + bx + c = 0 →
2x2 + 5x + 1 = 0
Recognize a = 2, b = 5, c = 1
Calculate the Discriminant
b2 – 4ac = (5)2 – 4(2)(1) = 25 – 8 = 17
There are two, real solutions

22. Example  Discriminant
Determine the nature of the solutions of:
0.3x2 − 0.4x = 0
SOLUTION
Recognize a = 0.3, b = −0.4, c = 0.8
Calculate the Discriminant
b2 − 4ac = (−0.4)2 − 4(0.3)(0.8) =0.16–0.96 = −0.8
Since the discriminant is negative, there are two NONreal complex-number solutions

23. Writing Equations from Solns
The principle of zero products informs that the this factored equation (x − 1)(x + 4) = 0 has solutions 1 and −4.
If we know the solutions of an equation, we can write an equation, using the principle in reverse.

24. Example  Write Eqn from solns
Find an eqn for which 5 & −4/3 are solns
SOLUTION
x = 5 or x = –4/3
x – 5 = 0 or x + 4/3 = 0
Get 0’s on one side
Using the principle of zero products
(x – 5)(x + 4/3) = 0
x2 – 5x + 4/3x – 20/3 = 0
Multiplying
3x2 – 11x – 20 = 0
Combining like terms and clearing fractions

25. Example  Write Eqn from solns
Find an eqn for which 3i & −3i are solns
SOLUTION
x = 3i or x = –3i
x – 3i = 0 or x + 3i = 0
Get 0’s on one side
Using the principle of zero products
(x – 3i)(x + 3i) = 0
x2 – 3ix + 3ix – 9i2 = 0
Multiplying
x = 0
Combining like terms

26. WhiteBoard Work Problems From §8.2 Exercise Set 18, 30, 44, 58
Solving Quadratic Equations
1. Check to see if it is in the form ax2 = p or (x + c)2 = d.
If it is, use the square root property
2. If it is not in the form of (1), write it in standard form:
ax2 + bx + c = 0 with a and b nonzero.
3. Then try factoring.
4. If it is not possible to factor or if factoring seems difficult, use the quadratic formula.
The solns of a quadratic eqn cannot always be found by factoring. They can always be found using the quadratic formula.

27. All Done for Today
The Quadratic Formula

28. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer –

29. Graph y = |x|
Make T-table

31. Quadratic Equation Graph
The graph of a quadratic eqn describes a “parabola” which has one of a:
Bowl shape
Dome shape
The graph, depending on the “Vertex” Location may have different numbers of x-intercepts: 2 (shown), 1, or NONE