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Math Help - Pre-Calculus Help!

  1. #1
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    Question Pre-Calculus Help!

    Need Help with :
    finding real solutions.
    x4-3x2+2=0
    the numbers after the x's are the powers.

    also if i need to type in square roots how do i do that?
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  2. #2
    Senior Member polymerase's Avatar
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    Quote Originally Posted by rsultana View Post
    Need Help with :
    finding real solutions.
    x4-3x2+2=0
    the numbers after the x's are the powers.

    also if i need to type in square roots how do i do that?
    <br />
(x^2-2)(x^2-1)<br />

    <br />
x=+\sqrt{2}, -\sqrt{2},1,-1<br />
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rsultana View Post
    Need Help with :
    finding real solutions.
    x4-3x2+2=0
    the numbers after the x's are the powers.

    also if i need to type in square roots how do i do that?
    ah, i see! you typed the question wrong last time, had me scratching my head
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    ah, i see! you typed the question wrong last time, had me scratching my head
    ooops i am so sorry...hey thanks for the help
    Finding real solutions:
    sqrt(3-sqrt(x+5))=2 <<<<< i dont know if i am writing this correct. let me know if you understand
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    Quote Originally Posted by polymerase View Post
    <br />
(x^2-2)(x^2-1)<br />

    <br />
x=+\sqrt{2}, -\sqrt{2},1,-1<br />
    can you please explain this to me. oh never mind i got it!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rsultana View Post
    can you please explain this to me.
    he rewrote x^4 - 3x^2 + 2 as \left( x^2 \right)^2 - 3 \left( x^2 \right) + 2, which is quadratic in x^2, (you can see this if you replaced x^2 with y for instance, you would get, y^2 - 3y + 2).

    so he just factored it as if it were a quadratic:

    you would get, (y - 2)(y - 1) = \left( x^2 - 2 \right) \left( x^2 - 1 \right)

    both of these last terms are the difference of two squares
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rsultana View Post
    ooops i am so sorry...hey thanks for the help
    Finding real solutions:
    sqrt(3-sqrt(x+5))=2 <<<<< i dont know if i am writing this correct. let me know if you understand
    \sqrt {3 - \sqrt {x + 5}} = 2 ........square both sides

    \Rightarrow 3 - \sqrt{x + 5} = 4

    \Rightarrow -\sqrt{x + 5} = 1 ..........sqaure both sides again

    \Rightarrow x + 5 = 1

    now continue

    remember to check your solutions at the end
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    \sqrt {3 - \sqrt {x + 5}} = 2 ........square both sides

    \Rightarrow 3 - \sqrt{x + 5} = 4

    \Rightarrow -\sqrt{x + 5} = 1 ..........sqaure both sides again

    \Rightarrow x + 5 = 1

    now continue

    remember to check your solutions at the end
    my solution manual says no real solution...why?
    oooooo...i got it...cos -.3166 is not equal to 2...is this correct?
    please explain this problem!
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rsultana View Post
    my solution manual says no real solution...why?
    oooooo...i got it...cos -.3166 is not equal to 2...is this correct?
    please explain this problem!
    Cosine? Where did that come from?

    No, it's because when you substitute x = -4 into the original equation, the equation reads that \sqrt{2} = 2, which is ridiculous.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rsultana View Post
    my solution manual says no real solution...why?
    oooooo...i got it...cos -.3166 is not equal to 2...is this correct?
    please explain this problem!
    this is why i said check your solution at the end. sometimes squaring both sides introduces erroneous solutions. if plugging in the answers you got makes no sense, or don't work, then either there's no solution, or you made a mistake
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Cosine? Where did that come from?
    hehe, i think the poster meant "'cause"
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    Quote Originally Posted by topsquark View Post
    Cosine? Where did that come from?

    No, it's because when you substitute x = -4 into the original equation, the equation reads that \sqrt{2} = 2, which is ridiculous.

    -Dan
    oops its not cosine it was supposed to be "because".....

    another question:
    inequality involving quotient: using interval notation, and sketch the solution
    2x-3/x+1 ≤1
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rsultana View Post
    oops its not cosine it was supposed to be "because".....

    another question:
    inequality involving quotient: using interval notation, and sketch the solution
    2x-3/x+1 ≤1
    First, new questions should be placed in new threads.

    BUT
    I already answered this in your other thread. If you cannot understand my solution that thread is the one you should make a mention of it. That's only polite.

    AND
    Please don't double post. See rule #1 here.

    -Dan
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  14. #14
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    Quote Originally Posted by topsquark View Post
    First, new questions should be placed in new threads.

    BUT
    I already answered this in your other thread. If you cannot understand my solution that thread is the one you should make a mention of it. That's only polite.

    AND
    Please don't double post. See rule #1 here.

    -Dan
    yes sir...i got it...thanks
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