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MATH 121, Lesson 2.6 Notes

by: Mallory McClurg

MATH 121, Lesson 2.6 Notes Math 121

Marketplace > University of Mississippi > Math > Math 121 > MATH 121 Lesson 2 6 Notes
Mallory McClurg
GPA 3.37

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About this Document

Examples of every type of radical equation we may need to solve, with step by step explanations!
College Algebra
Class Notes
radical, equation, Math, college, Algebra
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This 2 page Class Notes was uploaded by Mallory McClurg on Sunday September 11, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 23 views. For similar materials see College Algebra in Math at University of Mississippi.


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Date Created: 09/11/16
Math 121 Chapter 2 Notes Lesson 2.6 – Radical Equations Example 1. 3√(5z + 6z) = 3 (To solve this equation, we should first cub2 both 3 sides.) 5z + 6z = 3 (Simplify and then put into quadratic form.) 5z + 6z – 27 = 0 (Factor it out.) (5z – 9)(z + 3) = 0 (Set each group equal to zero and solve for x.) x = 9/5 x = -3 Example 2. √(5x + 19) + 8 = x + 7 (To begin, get the expression under the square root on a side by itself.) √(5x + 19) = x – 1 (Raise both sides to the power of 2.) 5x + 19 = (x – 1)(x – 1) (FOIL the groups on the right side of the equation.) 5x + 19 = x – 2x + 1 (Combine like terms!) 2 x – 7x – 18 = 0 (Factor it out.) (x + 2)(x – 9) = 0 (Solve for x.) x = -2 x = 9 (Verify both of these solutions in the original equation.) x ≠ -2 x = 9 (Only one solution!) Example 3. √(7z + 4) = √(5z – 1) + 1 (Raise both sides to the power of 2.) 7z + 4 = (√(5z – 1) + 1) 2 (To simplify the right side of the equation, use the formula (a + b) = a +2 2ab + b .) 2 7z + 4 = 5z – 1 + 2*√(5z – 1)*(1)+1 (Simplify this equation.) 7z + 4 = 2√(5z – 1) + 5z + 2 (Get the “2√(5z – 1)” on a side by itself.) 7z + 4 – 5z – 2 = 2√(5z – 1) (Combine like terms on the left side of the equation.) 2z + 2 = 2√(5z – 1) (To get rid of the square root symbol, square both sides of the equation.) (2z + 2) (2z + 2) = 2 * (5z – 1) (FOIL out the left side of the equation.) 2 4z + 8z + 4 = 20z + 4 (Combine like terms, so the equation is equal to zero.) 4z – 12z = 0 (Factor out the common “z”.) z (4z – 12) = 0 z = 0 z = 3 (Verify both solutions in the original equation; they both work!) 4xamp2e 4. 4 √(y + 9y) = √(3y + 27) (First we can raise both sides to the power of 4.) 2 y + 9y = 3y + 27 (Combine like terms on one side and set equal to zero.) y + 6y – 27 = 0 (Factor it out and solve for x.) (y + 9)(y – 3) = 0 y = -9 y = 3 (Verify. They both work!) Example 5. 3/2 y2/3– 64 = 0 (Add 64 to both sides.2/3 y = 64 (We should recognize that “y ” is the same thing as √(y ). Now, square both sides of the equation.) 3 y = 4096 (Take the cube root of both sides.) y = 16


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