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Radical Equations.

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Presentation on theme: "Radical Equations."— Presentation transcript:

1 Radical Equations

2 Review Simplify the given radical expression:

3 Radical equations - equations in which the unknown is included in a radicand. Example; 2 𝑥 + 5 = 9 2x + 5 = 9 – not a radical equation

4 Procedure to solve the radical equation
Isolate the term containing the radical. Square both sides Solve for the unknown. Check the roots obtained in the original equation.

5 Extraneous root - a value of the unknown which does not satisfy the original equation. It must be rejected. Example: 2𝑥 = -4 2x = 16 -squaring both sides x = 8 -dividing both sides by 2 *8 does not satisfy the original equation and must be rejected

6 ( 7𝑥+5 ) 2 = (3) 2 7𝑥+5=9 7𝑥=4 𝑥= 4 7 Examples:
Solve for the value of the unknown given the equation, 7𝑥+5 =3 ( 7𝑥+5 ) 2 = (3) 2 7𝑥+5=9 7𝑥=4 𝑥= 4 7

7 Example 2: 2. 5𝑥+3 = 3𝑥+7 𝑥=2

8 3. 𝑥 −3= 30 −2𝑥 𝑥=7, 𝑥 =−3(extraneous root)
Example 3: 3. 𝑥 −3= 30 −2𝑥 𝑥=7, 𝑥 =−3(extraneous root)

9 Example 4: 4. 3 𝑥 −1 +11=2𝑥 𝑥=10 , 𝑥=13/4

10 1. 17𝑥 − 𝑥 2 −5 =7 2. 𝑥 −1 −𝑥=−7 3. 2𝑥 −1 + 𝑥+4 =6 4. 5𝑥+4 −1=2𝑥
Exercises: 1. 17𝑥 − 𝑥 2 −5 =7 2. 𝑥 −1 −𝑥=−7 3. 2𝑥 −1 + 𝑥+4 =6 4. 5𝑥+4 −1=2𝑥

11 Problem Solving 1. Let ∆ABC as shown in the figure below. Let the height from C be Find BC. A B C 15 5 5 5

12 A B C 15 5 5 5 x (15) 2 = (5 5 ) 2 + (𝑥+5) 2 225= 𝑥 x + 25 𝑥 x -75 = 0 𝑥+15 𝑥 −5 =0 𝑥=−15, 𝑥=5 𝐵𝐶 2 = (5 5 ) 𝐵𝐶 2 =125+25 𝐵𝐶 2 =150 𝐵𝐶= 150 𝐵𝐶= 5 6

13 Problem 2 2. Two circles, A and B, have areas 176 and 44 meters squared respectively. What is the difference between the diameters of these two circles? 3. Two cylinders have the same shape. Cylinder A has a height of 7 feet and a volume of 6,600 feet cube. Cylinder B has a height of 14 feet and a volume of 22,000 feet cube. How many times is the radius of cylinder A compared to the radius of cylinder B.


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