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§ 8.2
The Quadratic Formula
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 8.2
The Quadratic Formula
The Quadratic FormulaThe solutions of a quadratic equation in standard form
with , are given by the quadratic formula
02 cbxax 0a
.2
42
a
acbbx
See page 576 of your textbook to see how the quadratic formula is derived using ‘completing the square’.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 8.2
The Quadratic Formula
EXAMPLEEXAMPLE
Solve using the quadratic formula:
SOLUTIONSOLUTION
The given equation is in standard form. Begin by identifying the values for a, b, and c.
.01582 xx
01582 xx
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
The Quadratic Formula
CONTINUECONTINUEDD
a
acbbx
2
42 Use the quadratic formula.
12
151488 2
xSubstitute the values for a, b, and c: a = 1, b = 8, c = 15.
2
60648 Simplify.
2
48 Subtract.
2
28 The square root of 4 is 2.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 8.2
The Quadratic Formula
CONTINUECONTINUEDD
2
28x
The solutions are -3 and -5. The solution set is {-3,-5}.
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.
2
28 x
32
6
x 5
2
10
x
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 8.2
The Quadratic Formula
EXAMPLEEXAMPLE
Solve using the quadratic formula:
SOLUTIONSOLUTION
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.
.542 2 xx
a = 2 b = 4 c = -5
542 2 xx This is the given equation.0542 2 xx Subtract -4x + 5 from
both sides.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 8.2
The Quadratic Formula
CONTINUECONTINUEDD
a
acbbx
2
42 Use the quadratic formula.
22
52444 2
xSubstitute the values for a, b, and c: a = 2, b = 4, c = -5.
4
40164 Simplify.
4
564 Add.
Substituting these values into the quadratic formula and simplifying gives the equation’s solutions.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 8.2
The Quadratic Formula
CONTINUECONTINUEDD
4
1424 14214414456
4
1422 Factor out 2 from the numerator.
2
142
Divide the numerator and denominator by 2.
The solutions are , and the solution set is2
142 .2
142
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
• The relationships among the various sets of numbers.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 8.2
Quadratic Formula – The Discriminant
The Discriminant and
the Kinds of Solutions to TDiscriminant 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
acb 42
02 cbxax
02 cbxax cbxaxy 2
042 acb
Page 580
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 8.2
Quadratic Formula – The Discriminant
The Discriminant and
the Kinds of Solutions to TDiscriminant 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
acb 42
02 cbxax
02 cbxax cbxaxy 2
042 acb
CONTINUECONTINUEDD
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 8.2
Quadratic Formula – The Discriminant
042 acb
CONTINUECONTINUEDD
The Discriminant and
the Kinds of Solutions to TDiscriminant Kinds of Solutions to Graph of
No real solution; two imaginary solutions;
The solutions are complex conjugates.
No x-intercepts
acb 42
02 cbxax
02 cbxax cbxaxy 2
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 8.2
Quadratic Formula – The Discriminant
EXAMPLEEXAMPLE
For each equation, compute the discriminant. Then determine the number and type of solutions:
SOLUTIONSOLUTION
.25204(b)0342(a) 22 xxxx
a = 2 b = -4 c = 3
Begin by identifying the values for a, b, and c in each equation. Then compute , the discriminant.acb 42
0342(a) 2 xx
Substitute and compute the discriminant:
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 8.2
Quadratic Formula – The Discriminant
25204(b) 2 xx
a = 4 b = -20 c = 25
.8241632444 22 acb
We must first put the quadratic equation in standard form.
CONTINUECONTINUEDD
The discriminant, -8, shows that there are two imaginary solutions. These solutions are complex conjugates of each other.
25204 2 xx
025204 2 xx Subtract 20x – 25 from both sides.
Substitute and compute the discriminant:
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 8.2
Quadratic Formula – The Discriminant
.04004002544204 22 acb
CONTINUECONTINUEDD
The discriminant, 0, shows that there is only one real solution. This real solution is a rational number.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 8.2
Quadratic Formula in Application
EXAMPLE: similar to # 79 in homeworkEXAMPLE: similar to # 79 in homework
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.
SOLUTIONSOLUTION
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.
x - 2
x
6
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 8.2
Quadratic Formula in Application
Now we can use the Pythagorean Theorem to create an equation that contains the information provided.
CONTINUECONTINUEDD
222 62 xx This is the Pythagorean Theorem.
36222 222 xxx Evaluate the exponents.
36442 2 xx Simplify.
03242 2 xx Subtract 36 from both sides.
01622 2 xx Factor 2 out of all terms on left side.
01622 xx 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 #18 Section 8.2
Quadratic Formula in Application
CONTINUECONTINUEDD
Substitute the values for a, b, and c into the quadratic formula.
12
161422 2
x
Simplify.2
6442 x
Simplify.2
682 x
Rewrite the radicand.2
1742 x
Rewrite as two radicals.2
1742 x
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 8.2
Quadratic Formula in Application
CONTINUECONTINUEDD
Simplify.2
1722 x
Factor 2 out of the numerator. 2
1712 x
Divide the numerator and denominator by 2.
171x
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.
feet 1.5171 x feet 1.3171 x
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 8.2
Quadratic Formula in Application
Check 6 on page 585Check 6 on page 585
The function
Models a woman’s normal systolic blood pressure, P(A), at age A. Use this function to find the age to the nearest year, of a woman whose normal systolic blood pressure is 115 mm Hg.
10705.001.0)( 2 AAAP
Determine a, b, and c to use the quadratic formula.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 8.2
Quadratic Formula in Application
CONTINUECONTINUEDD
Substitute the values for a, b, and c into the quadratic formula.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 8.2
Quadratic Formula in Application
CONTINUECONTINUEDD
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 8.2
Quadratic Formula in Application
Problem 84 on page 587: similar to #83 in Problem 84 on page 587: similar to #83 in homeworkhomework
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 8.2
Quadratic Formula in Application
Problem 84 on page 587, continuedProblem 84 on page 587, continued
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 8.2
Quadratic Formula in Application
Problem 84 on page 587, continuedProblem 84 on page 587, continued
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 8.2
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 astudy tip on this common error that many students make.
DONE
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 8.2
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 #29 Section 8.2
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.
2x
02 cbxax
cbxax 2
02 cax
0253 2 xx
0213 xx
02or 013 xx
2 3
1 xx
074 2 x74 2 x
4/72 x
2/72 x
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 8.2
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
; u is a first-degree polynomial.
Use the square root property.
du 2 54 2 x
CONTINUECONTINUEDD
54 x
54 x
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 8.2
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
cannot be factored or the factoring is too difficult.
Use the quadratic formula.
0622 xx
CONTINUECONTINUEDD
02 cbxaxcbxax 2
a = 1 b = -2 c = -6
12
61422 2
x
71x
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 8.2
The Zero-Product Principle
The Zero-Product Principle in ReverseIf A = 0 or B = 0, then AB = 0.
Blitzer, Intermediate Algebra, 5e – Slide #33 Section 8.2
The Zero-Product Principle
EXAMPLEEXAMPLE
Write a quadratic equation with the given solution set:
SOLUTIONSOLUTION
.3
1,
6
5
Because the solution set is , then
3
1,
6
5
3
1 or
6
5 xx
Obtain zero on one side of each equation.
06
5x 0
3
1x
Clear fractions, multiplying by 6 and 3 respectively.
056 x 013 x
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 8.2
The Zero-Product Principle
Use the zero-product principle in reverse.
01356 xx
CONTINUECONTINUEDD
Use the FOIL method to multiply.0515618 2 xxx
Combine like terms.05918 2 xx
Thus, one equation is . Many other quadratic
equations have for their solution sets.
05918 2 xx
3
1,
6
5