Solving radical equations

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  1. Solving Radical Equations Involving One Square Root Term
    1. 3x2+x=03\sqrt {x - 2} + x = 0
      (hint: no solutions)
    2. x2=2x+1x - 2 = 2\sqrt {x + 1}
      (hint: one solutions)
    3. x+2x27=3x + \sqrt {2{x^2} - 7} = 3
      (hint: two solutions)
    4. x+910=8\sqrt {x + 9} - 10 = - 8
      (hint: one solutions)
  2. Solving Radical Equations Involving More than One Square Root Term
    1. 3x+4x+1=3 \sqrt{3x+4}-\sqrt{x+1}=3
      (hint: one solutions)
    2. x2+3=2x+5\sqrt{x-2}+3=\sqrt{2x+5}
      (hint: two solutions)
    3. x4+x=4 \sqrt{x-4}+\sqrt{x}=4
      (hint: one solutions)
    4. 5x+20=x+8+x+3\sqrt{5x+20}=\sqrt{x+8}+\sqrt{x+3}
      (hint: two solutions)
Topic Notes
Radical equations are equations that have variables stunk inside a radical. We will show you how to solve this type of equations in this lesson.
3-step approach for solving radical equations:

1) Isolate the "square root" term.

2) i. Square both sides of the equation to get rid of the "square root".

ii. (optional) If the resulting equation still has a "square root" term, repeat steps 1 and 2.

3) Solve the equation, and REMEMBER to substitute answers back into the original equation to identify any EXTRANEOUS ROOTS,