1. § 8.2
The Quadratic Formula

2. The Quadratic Formula
We can use the method of completing the square to derive an equation that can be used to solve any quadratic equation – those that factor, and those that don’t.
This equation will enable you to solve equations more quickly than the method of completing the square. When quadratics are easy to factor, you will probably want to continue to use the method of factoring, for that will be quicker.
The formula that we will derive and use is called the quadratic formula. You will want to memorize this formula.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 8.2

3. The Quadratic Formula The Quadratic Formula
The solutions of a quadratic equation in standard form
with , are given by the quadratic formula
See page 575 of your textbook to see how the quadratic formula is derived using ‘completing the square’.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 8.2

4. The Quadratic Formula Solve using the quadratic formula:
EXAMPLE
Solve using the quadratic formula:
SOLUTION
The given equation is in standard form. Begin by identifying the values for a, b, and c.
a = 1
b = 8
c = 15
Substituting these values into the quadratic formula and simplifying gives the equation’s solutions.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 8.2

5. The Quadratic Formula Use the quadratic formula.
CONTINUED
Use the quadratic formula.
Substitute the values for a, b, and c: a = 1, b = 8, c = 15.
Simplify.
Subtract.
The square root of 4 is 2.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 8.2

6. The Quadratic Formula
CONTINUED
Now we will evaluate this expression in two different ways to obtain the two solutions. On the left, we will add 2 to -8. On the right, we will subtract 2 from -8.
The solutions are -3 and -5. The solution set is {-3,-5}.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 8.2

7. The Quadratic Formula Solve using the quadratic formula:
EXAMPLE
Solve using the quadratic formula:
SOLUTION
The quadratic equation must be in standard form to identify the values for a, b, and c. To move all terms to one side and obtain zero on the right, we subtract -4x + 5 from both sides. Then we can identify the values for a, b, and c.
This is the given equation.
Subtract -4x + 5 from both sides.
a = 2
b = 4
c = -5
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 8.2

8. The Quadratic Formula
CONTINUED
Substituting these values into the quadratic formula and simplifying gives the equation’s solutions.
Use the quadratic formula.
Substitute the values for a, b, and c: a = 2, b = 4, c = -5.
Simplify.
Add.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 8.2

9. The Quadratic Formula Factor out 2 from the numerator.
CONTINUED
Factor out 2 from the numerator.
Divide the numerator and denominator by 2.
The solutions are , and the solution set is
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 8.2

10. the Kinds of Solutions to T
Quadratic Formula – The Discriminant
The Discriminant and
the Kinds of Solutions to T
Discriminant
Kinds of Solutions to
Graph of
Two unequal real solutions
If a, b, and c are rational numbers and the discriminant is a perfect square, the solutions are rational. If the discriminant is not a perfect square, the solutions are irrational conjugates.
Two x-intercepts
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 8.2

11. the Kinds of Solutions to T
Quadratic Formula – The Discriminant
CONTINUED
The Discriminant and
the Kinds of Solutions to T
Discriminant
Kinds of Solutions to
Graph of
One real solution (a repeated solution);
If a, b, and c are rational numbers the repeated solution is also a rational number
One x-intercept
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 8.2

12. the Kinds of Solutions to T
Quadratic Formula – The Discriminant
CONTINUED
The Discriminant and
the Kinds of Solutions to T
Discriminant
Kinds of Solutions to
Graph of
No real solution; two imaginary solutions;
The solutions are complex conjugates.
No x-intercepts
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 8.2

13. Quadratic Formula – The Discriminant
EXAMPLE
For each equation, compute the discriminant. Then determine the number and type of solutions:
SOLUTION
Begin by identifying the values for a, b, and c in each equation. Then compute , the discriminant.
a = 2
b = -4
c = 3
Substitute and compute the discriminant:
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 8.2

14. Quadratic Formula – The Discriminant
CONTINUED
The discriminant, -8, shows that there are two imaginary solutions. These solutions are complex conjugates of each other.
We must first put the quadratic equation in standard form.
Subtract 20x – 25 from both sides.
a = 4
b = -20
c = 25
Substitute and compute the discriminant:
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 8.2

15. Quadratic Formula – The Discriminant
CONTINUED
The discriminant, 0, shows that there is only one real solution. This real solution is a rational number.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 8.2

16. Solving Quadratic Equations
Determining the Most Efficient Technique to Use When Solving a Quadratic Equation
Description and Form of the Quadratic Equation
Most Efficient Solution Method
Example
and
can be factored easily.
Factor and use the zero-product principle.
The quadratic equation has no x-term. (b = 0)
Solve for and apply the square root property.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 8.2

17. Solving Quadratic Equations
CONTINUED
Determining the Most Efficient Technique to Use When Solving a Quadratic Equation
Description and Form of the Quadratic Equation
Most Efficient Solution Method
Example
; u is a first-degree polynomial.
Use the square root property.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 8.2

18. Solving Quadratic Equations
CONTINUED
Determining the Most Efficient Technique to Use When Solving a Quadratic Equation
Description and Form of the Quadratic Equation
Most Efficient Solution Method
Example
and
cannot be factored or the factoring is too difficult.
Use the quadratic formula.
a = 1
b = -2
c = -6
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 8.2

19. The Zero-Product Principle in Reverse
If A = 0 or B = 0, then AB = 0.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 8.2

20. The Zero-Product Principle
EXAMPLE
Write a quadratic equation with the given solution set:
SOLUTION
Because the solution set is , then
Obtain zero on one side of each equation.
Clear fractions, multiplying by 6 and 3 respectively.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 8.2

21. The Zero-Product Principle
CONTINUED
Use the zero-product principle in reverse.
Use the FOIL method to multiply.
Combine like terms.
Thus, one equation is Many other quadratic
equations have for their solution sets.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 8.2

22. Quadratic Formula in Application
EXAMPLE
The hypotenuse of a right triangle is 6 feet long. One leg is 2 feet shorter than the other. Find the lengths of the legs. Round to the nearest tenth of a foot.
SOLUTION
Since the hypotenuse is 6 feet long, and one leg of the triangle, x, is 2 feet longer than the other leg, x - 2, the triangle can be represented as follows. 6
x - 2 x
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 8.2

23. Quadratic Formula in Application
CONTINUED
Now we can use the Pythagorean Theorem to create an equation that contains the information provided.
This is the Pythagorean Theorem.
Evaluate the exponents.
Simplify.
Subtract 36 from both sides.
Factor 2 out of all terms on left side.
Divide both sides by 2.
Determine a, b, and c to use the quadratic formula.
a = 1
b = -2
c = -16
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 8.2

24. Quadratic Formula in Application
CONTINUED
Substitute the values for a, b, and c into the quadratic formula.
Simplify.
Simplify.
Rewrite the radicand.
Rewrite as two radicals.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 8.2

25. Quadratic Formula in Application
CONTINUED
Simplify.
Factor 2 out of the numerator.
Divide the numerator and denominator by 2.
Now and to the nearest tenth of a foot. The answer -3.1 feet is of course impossible. Therefore, the length of the side labeled x must be 5.1 feet. Therefore, the side labeled x – 2 must be 5.1 – 2 = 3.1 feet.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 8.2

26. In Conclusion – an error to watch for
Note:
Many students use the quadratic formula correctly until the last step, where they make an error in simplifying the solutions. Be sure that you factor the numerator before dividing the numerator and denominator by the greatest common factor. Remember. you can only cancel factors of the whole numerator.
You cannot divide just one term in the numerator and denominator by their greatest common factor. See page 578 in your text for a
study tip on this common error that many students make.
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 8.2