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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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For more information, please see full course syllabus of Algebra 1
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Lecture Comments (4)

1 answer

Last reply by: Dr Carleen Eaton
Sat Jul 27, 2013 10:33 AM

Post by Tami Cummins on July 18, 2013

In 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 answer

Last reply by: Dr Carleen Eaton
Fri May 24, 2013 9:43 PM

Post by Anwar Alasmari on May 23, 2013

In 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?

Solving Radical Equations

  • A radical equation is an equation that contains radicals with variables in the radicand.
  • 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 solution is 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.

Solving Radical Equations

√{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.

Answer

Solving Radical Equations

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
  • Radical Equations 0:15
    • Examples
  • Solving a Radical Equation 1:13
    • Isolate Radical
    • 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