INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

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 1 answerLast reply by: Dr Carleen EatonSat Jul 27, 2013 10:33 AMPost by Tami Cummins on July 18, 2013In example 3 when you are finding the difference of two squares shouldn't the 6 be positive or a plus 6 due to minus 2ab and b is a negative 3? 1 answerLast reply by: Dr Carleen EatonFri May 24, 2013 9:43 PMPost by Anwar Alasmari on May 23, 2013In example II, radical 9 has to be 3 and -3 because (3)(3)=9 as well as (-3)(-3) also equals 9. Is this right?

• To solve a radical equation, isolate the radical on one side of the equation and then square both sides of the equation. This will eliminate at least one radical. If a radical remains in the new equation, repeat the process.
• An extraneous solutionis a value which produces a negative radicand in one or more of the radicals in the original equation. Check all potential solutions in the original equation. Exclude extraneous values from the solution set.

√{4x − 16} = 3√2
• ( √{4x − 16} )2 = ( 3√2 )2
• 4x − 16 = 9 ×2
• 4x − 16 = 18
• 4x = 34
x = 8[1/2]
√{10x + 12} = 8√3
• 10x + 12 = 64 ×3
• 10x + 12 = 192
• 10x = 180
x = 18
4√5 = √{6y + 8}
• 16 ×5 = 6y + 8
• 80 = 6y + 8
• 72 = 6y
12 = y
x = √{x + 30}
• x2 = ( √{x + 30} )2
• x2 = x + 30
• x2 − x − 30 = 0
• ( x + 5 )( x − 6 ) = 0
x = − 5,6
x = √{x + 56}
• x2 = ( √{x + 56} )2
• x2 = x + 56
• x2 − x − 56 = 0
• ( x + 7 )( x − 8 ) = 0
x = − 7,8
x = √{13x + 30}
• x2 = ( √{13x + 30} )2
• x2 = 13x + 30
• x2 − 13x − 30 = 0
• ( x + 2 )( x − 15 ) = 0
x = − 2,15
− 2 + √{x + 4} = x + 4
• √{x + 4} = x + 4 + 2
• √{x + 4} = x + 6
• ( √{x + 4} )2 = ( x + 6 )2
• x + 4 = x2 + 12x + 36
• 4 = x2 + 11x + 36
• 0 = x2 + 11x + 28
• x2 + 11x + 28 = 0
• ( x + 4 )( x + 7 ) = 0
x = − 4, − 7
1 + √{4x − 11} = x − 1
• √{4x − 11} = x − 1 − 1
• √{4x − 11} = x − 2
• ( √{4x − 11} )2 = ( x − 2 )2
• 4x − 11 = x2 − 4x + 4
• − 11 = x2 − 8x + 4
• 0 = x2 − 8x + 15
• x2 − 8x + 15 = 0
• ( x − 5 )( x − 3 )
x = 5,3
x − √{3 − 11x} = 3
• − √{3 − 11x} = 3 − x
• ( − √{3 − 11x} )2 = ( 3 − x )2
• 3 − 11x = ( − x + 3 )2
• 3 − 11x = x2 − 6x + 9
• − 11x = x2 − 6x + 6
• 0 = x2 + 5x + 6
• 0 = ( x + 3 )( x + 2 )
x = − 3, − 2
x − √{5 − 7x} = 7
• − √{5 − 7x} = 7 − x
• ( − √{5 − 7x} )2 = ( − x + 7 )2
• 5 − 7x = x2 − 14x + 49
• − 7x = x2 − 14x + 44
• 0 = x2 − 7x + 44
• 0 = ( x − 11 )( x + 4 )
x = 11, − 4

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Examples
• Solving a Radical Equation 1:13
• Square Both Sides
• Example
• Extraneous Solutions 2:57
• Example: Check Solutions
• Example 1: Solve Equation 6:29
• Example 2: Solve Equation 9:52
• Example 3: Solve Equation 14:29
• Example 4: Solve Equation 20:53