1. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 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. Previously we solved 2 nd 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). Factor left hand side then apply the zero product rule. Solution: Square Root Property

2. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 2 Solution: or 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. Your Turn Problem #1 Recall the two methods for simplifying a radical. or 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. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 3 In this example, we will have a binomial squared on the LHS. We will still be able to use the square root property. Solution: Use the square root property. Don’t forget the ± on the RHS. Then simplify and solve for x. +3 Usually the radical part is written at the end of the expression. Your Turn Problem #2

4. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 4 Your Turn Problem #3 Solution: 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.

5. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 5 Solution: Your Turn Problem #4

6. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 6 In this example, we will have a binomial squared on the LHS. We will still be able to use the square root property. Solution: Use the square root property. Don’t forget the ± on the RHS. Then simplify both sides and solve for x. +3 Usually the radical part is written at the end of the expression. Your Turn Problem #5

7. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 7 Solution: Use the square root property. Simplify both sides, then subtract 1 on both sides. 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. 2 2 2 Your Turn Problem #6

8. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 8 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 22 2 Your Turn Problem #7

9. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 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 a c The Pythagorean Theorem is simply: Example: 5 12 13 This theorem holds true for the given triangle. 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. 7.2 Quadratic Equations and the Square Root Property BobsMathClass.Com Copyright © 2010 All Rights Reserved. 10 We can now make use of the Pythagorean Theorem to solve for d. 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. 5 ft Your Turn Problem #8 11 ft c=? The End. B.R. 6-10-07