7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 A second degree equation in one variable

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7.2 Quadratic Equations and the Square Root Property

BobsMathClass.Com Copyright © 2010 All Rights Reserved.

A second degree equation in one variable is an equation that contains the variable with an exponent of 2, but no higher power. Such equations are called quadratic equations.

2Any equation that can be written in the form is a quadratic equation.

Here a, b and c are real ax + bx + c =

numbers0

, a

0. 2 2Some examples of quadratic equations are: x 2x 48 0 and 3n 2n 1 0.

Previously we solved 2nd degree equations (the word “quadratic” was not used at that time) by factoring and applying the property: If ab = 0, then a= 0 or b = 0. (Principle of zero products). 2 Solve n 8n 48 0 by fExampl actore: ing.

Factor left hand side then apply the zero product rule.

Solution: n 12 n 4 0

n 12 0 or n 4 0

n 12 or n 4

4,12

2If x 16, then what numbers would satisfy this equation? Well 4 would work but so would 4.We could get those numbers by taking the square root of 16 and then placing both plus and minusin front of the answer. We will now state this property.

Square Root Property

2For any real number a, if x a , then x = a or x a

x = a or x = a can be written as x = a

2



2 SolExampl ve e . x 1 32

Solution: x 32 Using the Square Root Property

x 4 2 Simplify the radica *l.

The solution set is 4 2 32 16 2 *

4 2

or 32 2 2 2 2 2

2 2 24 2

Circle the pairs. For every pair circled, one of the numbers will go in front of the radical. Any numbers not circled will stay inside the radical.

2Solve using the square root property: x 60

Your Turn Problem #1

Answer: 2 15

Recall the two methods for simplifying a radical.

60 4 15

2 15

or 60 2 2 3 5

2 3 5

2 15

Circle the pairs. For every pair circled, one of the numbers will go in front of the radical. Any numbers not circled will stay inside the radical.

3



x 3 2 6

In this example, we will have a binomial squared on the LHS. We will still be able to use the square root property.

2 Solve Example 2. x 3 24

Solution:Use the square root property. Don’t forget the ± on the RHS. Then simplify and solve for x.

x 3 24

+3 +3

x 3 2 6 Usually the radical part is written at the end of the expression.

The solution set is 3 2 6

2Solve using the square root property: x+2 80


Answer: 2 4 5

4



2Solve using the square root property: x 75


2 SolveExample 3 x. 72

Solution: x 72 Using the Square Root Property.

x i 72 To get the radicand positive.

The solution set is 6i 2

x 6i 2 Simplifying the radical.

Answer: 5i 3

Recall the process of simplifying the square root of a fraction where we need to rationalize the denominator. We will need this information for the next example.

12Simplify: 7

a bMultiply by , then simplify each.b b

77

8449

2 217

84 2 2 3 7

2 21

5



2 SolExampl ve e . 5x 4 8

Solution: 2 28x We need to divide by 5 on both sides to get the x by itself.

5

8x Using the square root property.

5

8 5x Rationalize the denominator and simplify.55

2 10The solution set is .5

40 2 10x

525

2Solve using the square root property: 7x 18


3 14Answer:

7

6



x 3 2 6

In this example, we will have a binomial squared on the LHS. We will still be able to use the square root property.

2 Solve xExample . 3 5 24

Solution:

Use the square root property. Don’t forget the ± on the RHS. Then simplify both sides and solve for x. x 3 24

+3 +3

x 3 2 6 Usually the radical part is written at the end of the expression.

The solution set is 3 2 6 .

2Solve using the square root property: x+2 80


Answer: 2 4 5

7



2x 1 3i 5

2 Solve 2xExample +1 6. 45

Solution: Use the square root property. 2x 1 45

Simplify both sides, then subtract 1 on both sides. -1 -1

2x 1 3 5i Divide by 2 on both sides to get x by itself on the LHS. Anytime we have a complex number, it must be written in standard form, a +bi.

1 3 5The solution set is i2 2

2 2 2

2Solve using the square root property: 3x 2 27


2Answer: 3i

3

8



2Solve 2 2xExample 7. 3 4 24

Solution:

Now we can use the square root property.

Our first step is to isolate the squared binomial on the LHS before we can use the square root property -4 -4

22 2x 3 4 24

22 2x 3 20

2 22 22x 3 10

2x 3 10

2x 3 10

3 10The solution set is 23 10

x2

2Solve using the square root property: 3 4x 1 1 17

Your Turn Problem #71 6

Answer: i4 4

9



Pythagorean Theorem

Observe the right triangle to the right.A right triangle is a triangle where one of its angles is 90°. a and b are called legs of the triangle and c is called the hypotenuse.

b

ac

The Pythagorean Theorem is simply:2 2 2a b c

Example:

5

1213

This theorem holds true for the given triangle.

2 2 2

2 2 2

Using a b c , 5 12 13 25 144 169 169 169

ü

Next Slide

We will now use the Pythagorean Theorem and the Square Root Property to find an unknown side. Note that when we use the Square Root Property, we place a ± symbol on the RHS. Since the length of a side can not be negative, we will not use the ± symbol.

10



We can now make use of the Pythagorean Theorem to solve for d.

2 2 2d 12 16 2d 144 256 2d 112

d 112 4 7 10.6

Example 8. A 16-foot ladder resting against a house reaches a window-sill 12 feet above the ground. How far is the foot of the ladder from the foundation of the house? Express your answer in simplest radical form and to the nearest tenth of a foot.

Solution:

d = ?

Ladder

16 ft Wall 12ft

It would be appropriate to sketch the triangle made with the given information. d is the distance from the from the foot of the ladder to the foundation of the house.

The distance from the foot of the ladder tothe foundation is 4 7 or approximately 10.6 ft.

5 ft

Given the following right triangle, find the length of the missing side. Approximate to the nearest tenth.


11 ft

c=?

Answer: c 12.1 ft.

The End.B.R.6-10-07

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7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 A second degree equation in one variable