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anonymous

  • 5 years ago

find a radical equation of the form sqrt(ax+b)=x+c so that one solution is extraneous

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  1. anonymous
    • 5 years ago
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    okay. give an example of square root of a number.

  2. anonymous
    • 5 years ago
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    4

  3. anonymous
    • 5 years ago
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    okay so 4 is the square root of 16. so ax+b =16 and x+c = 4.

  4. anonymous
    • 5 years ago
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    say x is 3, then c is 1. and 3a+b is 16. lets say a is 4, then b is also 4.

  5. anonymous
    • 5 years ago
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    so your equation could be sqrt(4x+4)=x+1

  6. anonymous
    • 5 years ago
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    to find x, square both sides. and post what you got.

  7. anonymous
    • 5 years ago
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    -x^2+4x+3=0

  8. anonymous
    • 5 years ago
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    x=-1,-3

  9. anonymous
    • 5 years ago
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    wait. what is (x+1)^2?

  10. anonymous
    • 5 years ago
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    x^2+1

  11. anonymous
    • 5 years ago
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    (a+b)^2 is a^2+b^2? are you sure about that?

  12. anonymous
    • 5 years ago
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    its 4x+4=x^2+1^2

  13. anonymous
    • 5 years ago
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    do you know the formula for \[(a+b)^{2} \]

  14. anonymous
    • 5 years ago
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    okay, if you don't know the formula, you can tell me. I will explain.

  15. anonymous
    • 5 years ago
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    wait i no this its 4x+4=x^2+2x+1

  16. anonymous
    • 5 years ago
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    yes! solve that expression then.

  17. anonymous
    • 5 years ago
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    (x=-1,3)

  18. anonymous
    • 5 years ago
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    y, plug in x = 3 and x = -1 in your original expression, sqrt(4x+4)=x+1

  19. anonymous
    • 5 years ago
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    oh wait. this is a bad example.

  20. anonymous
    • 5 years ago
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    well, anyway, you get the idea. you have to follow the procedure I outlines, such that the original expression is solve by one value of x and the square expression is solved by 2 values of x, giving you one spurious value of x.

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