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anonymous

  • 5 years ago

Factoring a square root... sqr(x-2)-sqr(2x+6)-1 I need to turn this into a quadratic equation.

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  1. anonymous
    • 5 years ago
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    \[\sqrt{x-2}-\sqrt{2x+6}-1\]

  2. anonymous
    • 5 years ago
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    Square everything and solve normally.

  3. anonymous
    • 5 years ago
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    ACK! That was so simple... I don't know why I didn't remember that!!!!

  4. anonymous
    • 5 years ago
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    Now would the negative signs turn to positives?

  5. anonymous
    • 5 years ago
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    I mean the central negative sign and the sign in front of 1

  6. anonymous
    • 5 years ago
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    Nope. All of the quantities are squared. Like, each of the roots and it's -(1^2) for the constant.

  7. anonymous
    • 5 years ago
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    Okay great, that helped out so much!

  8. anonymous
    • 5 years ago
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    So just to be sure, x=-1?

  9. anonymous
    • 5 years ago
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    Another way to think of it would be to write out the exponents of each quantity, like the root is to the one half, then if you multiply (1/2)(2), you get (2/2)=1 right? So the exponents cancel but the quantity inside stays the same.

  10. anonymous
    • 5 years ago
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    No, it doesn't equal -1, I missed an addition step.

  11. anonymous
    • 5 years ago
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    The result would be something like -x+3

  12. anonymous
    • 5 years ago
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    x=3

  13. anonymous
    • 5 years ago
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    You get -x+3 because if you square everything, you get x-2-x+6-1, so combine like terms.

  14. anonymous
    • 5 years ago
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    It seems right until I plut it into \[\sqrt{2(3)+6}\] That turns into:\[\sqrt{12}\] But there is no complete square root of 12, so whatever it is -1, it can't be right.

  15. anonymous
    • 5 years ago
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    That leaves me with: 1-\[1-\sqrt{12}-1=0\]

  16. anonymous
    • 5 years ago
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    No, that's not it. The result is not a number unless you have something to set it equal to. You get an expression as an answer.

  17. anonymous
    • 5 years ago
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    In the original problem, the -1 was =1 and I moved it to the left side, I just didn't include that step.

  18. anonymous
    • 5 years ago
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    What was the left side originally?

  19. anonymous
    • 5 years ago
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    \[\sqrt{x-2}-\sqrt{2x+6}=1\]

  20. anonymous
    • 5 years ago
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    Either way, if I square the 1 on either side, I still end up with -1.

  21. anonymous
    • 5 years ago
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    No, I mean the full equation. In your question, you gave an expression. An equation must have an equal sign. Was zero on the other side of the equation originally? Or was nothing there?

  22. anonymous
    • 5 years ago
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    The last equation I posted was the full equation.

  23. anonymous
    • 5 years ago
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    Then your answer will be an expression, and you cannot move it to the other side.

  24. anonymous
    • 5 years ago
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    It says I have to solve for x though, I have to create 0 to solve for x.

  25. anonymous
    • 5 years ago
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    Oh so you mean it was given to you already set equal to one? In the problem they gave you?

  26. anonymous
    • 5 years ago
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    Yes, exactly. I just left out that step in my original post. The intent was to save you some work, but I will know better next time :P

  27. anonymous
    • 5 years ago
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    Lol well you have to say that you need to solve for x, otherwise the answer will be an expression. Then here's what you do: \[\sqrt{x-2} - \sqrt{x+6} = 1\] Which is the same as: \[(x-2)^{1/2} - (x+6)^{1/2} = 1\] In order to solve, we must get rid of the root, and we do this by squaring the entire equation. \[(x-2)^{(1/2)(2)} - (x+6)^{(1/2)(2)} = 1^{2}. = x - 2 - x + 6 = 1\]

  28. anonymous
    • 5 years ago
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    Oh! I tried to do that at first, but I didn't think to square those answers.

  29. anonymous
    • 5 years ago
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    Yeah, just do all that, and then solve for x like a normal linear equation.

  30. anonymous
    • 5 years ago
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    Thank you, ahhhh, I was so close :P

  31. anonymous
    • 5 years ago
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    Haha no worries. And I just realized I forgot the 2x so just pretend it's there lol

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