Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant 9-9 The Quadratic Formula and the Discriminant Holt Algebra 1 Warm Up Warm Up Lesson Presentation.

Holt Algebra 1

9-9 The Quadratic Formula and the Discriminant9-9 The Quadratic Formula

and the Discriminant

Holt Algebra 1

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 1

9-9 The Quadratic Formula and the Discriminant

Warmup

Solve using square roots. Check your answers.

1. x2 – 195 = 1

2. 4x2 – 18 = –9

3. 2x2 – 10 = –12

4. Solve 0 = –5x2 + 225. Round to the nearest hundredth.

± 14

± 6.71

no real solutions

Holt Algebra 1


Lesson Quiz: Part II

1. A community swimming pool is in the shape of a trapezoid. The height of the trapezoid is twice as long as the shorter base and the longer base is twice as long as the height.

The area of the pool is 3675 square feet. What is the length of the longer base? Round to the nearest foot.

(Hint: Use )

108 feet

Holt Algebra 1


Warm Up

Evaluate for x =–2, y = 3, and z = –1.

6 1. x2 2. xyz

3. x2 – yz 4. y – xz

4

5. –x 6. z2 – xy

7 1

7 2

Holt Algebra 1


Solve quadratic equations by using the Quadratic Formula.

Determine the number of solutions of a quadratic equation by using the discriminant.

Objectives

Holt Algebra 1


discriminant

Vocabulary

Holt Algebra 1


In the previous lesson, you completed the square to solve quadratic equations. If you complete the square of ax2 + bx + c = 0, you can derive the Quadratic Formula. The Quadratic Formula is the only method that can be used to solve any quadratic equation.

Holt Algebra 1


Holt Algebra 1


Holt Algebra 1


Holt Algebra 1


To add fractions, you need a common denominator.

Remember!

Holt Algebra 1

9-9 The Quadratic Formula and the DiscriminantExample 1A: Using the Quadratic Formula

Solve using the Quadratic Formula.

6x2 + 5x – 4 = 0

6x2 + 5x + (–4) = 0 Identify a, b, and c.

Use the Quadratic Formula.

Simplify.

Substitute 6 for a, 5 for b, and –4 for c.

Holt Algebra 1


Example 1A Continued


6x2 + 5x – 4 = 0

Simplify.

Write as two equations.

Solve each equation.

Holt Algebra 1

9-9 The Quadratic Formula and the DiscriminantExample 1B: Using the Quadratic Formula


x2 = x + 20

1x2 + (–1x) + (–20) = 0 Write in standard form. Identify a, b, and c.

Use the quadratic formula.

Simplify.

Substitute 1 for a, –1 for b, and –20 for c.

Holt Algebra 1


Example 1B Continued


x = 5 or x = –4

Simplify.



x2 = x + 20

Holt Algebra 1


You can graph the related quadratic function to see if your solutions are reasonable.

Helpful Hint

Holt Algebra 1


Many quadratic equations can be solved by graphing, factoring, taking the square root, or completing the square. Some cannot be solved by any of these methods, but you can always use the Quadratic Formula to solve any quadratic equation.

Holt Algebra 1


Example 2: Using the Quadratic Formula to Estimate Solutions

Solve x2 + 3x – 7 = 0 using the Quadratic Formula.

Use a calculator: x ≈ 1.54 or x ≈ –4.54.

Check reasonableness

Holt Algebra 1


If the quadratic equation is in standard form, the discriminant of a quadratic equation is b2 – 4ac, the part of the equation under the radical sign. Recall that quadratic equations can have two, one, or no real solutions. You can determine the number of solutions of a quadratic equation by evaluating its discriminant.

Holt Algebra 1


Holt Algebra 1


Holt Algebra 1


3x2 – 2x + 2 = 0 2x2 + 11x + 12 = 0 x2 + 8x + 16 = 0

a = 3, b = –2, c = 2 a = 2, b = 11, c = 12 a = 1, b = 8, c = 16

b2 – 4ac b2 – 4ac b2 – 4ac

(–2)2 – 4(3)(2) 112 – 4(2)(12) 82 – 4(1)(16)

4 – 24 121 – 96 64 – 64

–20 25 0

b2 – 4ac is negative.

There are no real solutions

b2 – 4ac is positive.

There are two real solutions

b2 – 4ac is zero.

There is one real solution

Example 3: Using the DiscriminantFind the number of solutions of each equation using the discriminant.

A. B. C.

Holt Algebra 1


The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = –16t2 + vt + c, where c is the beginning height of the object above the ground.

Holt Algebra 1


If the object is shot straight up from the ground, the initial height of the object above the ground equals 0.

Helpful Hint

Holt Algebra 1


The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = –16t2 + vt + c, where c is the initial height of the object above the ground. The ringer on a carnival strength test is 2 feet off the ground and is shot upward with an initial velocity of 30 feet per second. Will it reach a height of 20 feet? Use the discriminant to explain your answer.

Example 4: Application

Holt Algebra 1


Example 4 Continued

h = –16t2 + vt + c

20 = –16t2 + 30t + 2

0 = –16t2 + 30t + (–18)

b2 – 4ac

302 – 4(–16)(–18) = –252

Substitute 20 for h, 30 for v, and 2 for c.

Subtract 20 from both sides.

Evaluate the discriminant.

Substitute –16 for a, 30 for b, and –18 for c.

The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.

Holt Algebra 1


There is no one correct way to solve a quadratic equation. Many quadratic equations can be solved using several different methods.

Holt Algebra 1

9-9 The Quadratic Formula and the DiscriminantExample 5: Solving Using Different Methods

Solve x2 – 9x + 20 = 0. Show your work.

Method 1 Solve by graphing.

y = x2 – 9x + 20Write the related quadratic

function and graph it.

The solutions are the x-intercepts, 4 and 5.

Holt Algebra 1



Method 2 Solve by factoring.

x2 – 9x + 20 = 0

Example 5 Continued

(x – 5)(x – 4) = 0

x – 5 = 0 or x – 4 = 0

x = 5 or x = 4

Factor.

Use the Zero Product Property.


Holt Algebra 1



Method 3 Solve by completing the square.

Example 5 Continued

x2 – 9x + 20 = 0

Add to both sides.

x2 – 9x = –20

x2 – 9x + = –20 +

Factor and simplify.

Take the square root of both sides.

Holt Algebra 1



Method 3 Solve by completing the square.

Example 5 Continued


x = 5 or x = 4

Holt Algebra 1


Method 4 Solve using the Quadratic Formula.

Example 5: Solving Using Different Methods.

1x2 – 9x + 20 = 0

x = 5 or x = 4

Identify a, b, c.

Substitute 1 for a, –9 for b, and 20 for c.

Simplify.




Holt Algebra 1


Notice that all of the methods in Example 5 (pp. 655-656) produce the same solutions, –1 and –6. The only method you cannot use to solve x2 + 7x + 6 = 0 is using square roots. Sometimes one method is better for solving certain types of equations. The following table gives some advantages and disadvantages of the different methods.

Holt Algebra 1


Holt Algebra 1


Quadratic Formula

Common Errors

Holt Algebra 1


Common Quadratic Formula Errors7x2 – 20x + 9

a = 7 b = -20 c = 9

Error #1: Entering b2 into the calculator

-202 = -400instead of +400

Correction:You should enter (-20)2 = 400

or just 202

Holt Algebra 1


Common Quadratic Formula Errors

7x2 – 20x + 9

a = 7 b = -20 c = 9

Error #2:

Forgetting to take the square

root of the discriminate.

Holt Algebra 1


Common Quadratic Formula Errors

7x2 – 20x + 9

a = 7 b = -20 c = 9

Error #3: Entering –b for negative values of b.

- (-20) = +20

Holt Algebra 1

9-9 The Quadratic Formula and the DiscriminantCommon Quadratic Formula Errors

7x2 – 20x + 9

a = 7 b = -20 c = 9

Error #4: Forgetting that your calculator knows orders of operation and does exactly what you tell it to do.

6+4/2 = 8 Correct Answer = 5

Correction:

You should enter (6+4)/2 or

6+4, ENTER, Ans/2

2

46

Holt Algebra 1


L9-9 NOTES:

Holt Algebra 1


ASSIGNMENT:

• L9-9 pg 657 #9-69x3, skip #21, 54,

add #23, 53

#63 use another method

Holt Algebra 1


WARMUP

1. Solve –3x2 + 5x = 1 by using the Quadratic Formula.

2. Find the number of solutions of 5x2 – 10x – 8 = 0 by using the discriminant.

3. The height h in feet of an object shot straight up is modeled by h = –16t2 + vt + c, where c is the beginning height of the object above the ground. An object is shot up from 4 feet off the ground with an initial velocity of 48 feet per second. Will it reach a height of 40 feet? Use the discriminant to explain your answer.

4. Solve 8x2 – 13x – 6 = 0. Show your work.

Holt Algebra 1


Lesson Quiz: Part I 1. Solve –3x2 + 5x = 1 by using the Quadratic

Formula.

2. Find the number of solutions of 5x2 – 10x – 8 = 0 by using the discriminant.

≈ 0.23, ≈ 1.43

2

Holt Algebra 1


Lesson Quiz: Part II

The discriminant is zero. The object will reach its maximum height of 40 feet once.

4. Solve 8x2 – 13x – 6 = 0. Show your work.

3. The height h in feet of an object shot straight up is modeled by h = –16t2 + vt + c, where c is the beginning height of the object above the ground. An object is shot up from 4 feet off the ground with an initial velocity of 48 feet per second. Will it reach a height of 40 feet? Use the discriminant to explain your answer.

Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant 9-9 The Quadratic Formula and the Discriminant Holt Algebra 1 Warm Up Warm Up Lesson Presentation.

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