Radical Equations

Review Simplify the given radical expression: 2 3 +3 2 5 2 +2 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

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.

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

( 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

Example 2: 2. 5𝑥+3 = 3𝑥+7 𝑥=2

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

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

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

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

A B C 15 5 5 5 x (15) 2 = (5 5 ) 2 + (𝑥+5) 2 225=125 + 𝑥 2 + 10x + 25 𝑥 2 + 10x -75 = 0 𝑥+15 𝑥 −5 =0 𝑥=−15, 𝑥=5 𝐵𝐶 2 = (5 5 ) 2 + 5 2 𝐵𝐶 2 =125+25 𝐵𝐶 2 =150 𝐵𝐶= 150 𝐵𝐶= 5 6

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.