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Grazes

  • 2 years ago

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  1. Grazes
    • 2 years ago
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  2. yummydum
    • 2 years ago
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    square everything to get rid of the radicals

  3. yummydum
    • 2 years ago
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    then solve for x after combining like terms

  4. Azteck
    • 2 years ago
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    It's pretty dangerous when you're squaring in mathematics. Because you create new values. But in this case, I think it's okay. When you get to calculus, you wouldn't want to square stuff.

  5. tkhunny
    • 2 years ago
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    If you keep track, there is little risk. 2x - 3 >= 0 leads to x >= 3/2 x + 2 >= 0 leads to x >= -2 Therefore, if we promise to keep x >= 3/2, we generally can do as we wish. The second methodology would be to check final potential answers in ORIGINAL equations if they exist. Please don't just square in the present condition. First, isolate one of the radicals. There are other failsafes that we can track along the way. \(\sqrt{2x-3} = 1 + \sqrt{x+2}\). Everything is positive, here. There should be little or no risk in the squaring. \(2x - 3 = 1 + 2\sqrt{x+2} + (x+2)\) Simplify, Isolate, and square it again. ALWAYS keep your eyes on the Domain issues.

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