Page 1 of 2 12 LastLast
Results 1 to 15 of 19

Math Help - Help me please

  1. #1
    Junior Member
    Joined
    Jul 2006
    Posts
    25

    Help me please

    I just dont get it. Can someone please help.


    Problems are attached


    Chester
    Last edited by Chester; August 2nd 2006 at 06:21 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    1. \frac{b}{3}=\frac{-3}{4} therefore 4b=-9 then b=-\frac{9}{4}

    Find each root: your answer is correct.
    Last edited by Quick; August 1st 2006 at 03:07 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    These should be explained to you better, so here I go:

    If you have an equation like: \frac{a}{b}=\frac{x}{y} than you can use the cross-product rule to say: ay=bx

    Radicals: Just something most people tend to not know, \sqrt[x]{a}=(a)^{\frac{1}{x}}

    therefore when you see something like \sqrt[3]{5b^9} you can say it equals (5b^9)^{\frac{1}{3}} and then use the rule of exponents to say: 5^{\frac{1}{3}}\cdot b^{\frac{9}{3}} which simplifies to b^3\sqrt[3]{5}

    for some reason I really don't feel like answer most of your questions now, so I'll let someone else do that.
    Last edited by Quick; August 1st 2006 at 02:50 PM. Reason: fixing cross product rule
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2006
    Posts
    25

    Thanks

    Quick,
    Thank you very much for the explanation. Any help is very much appreciated.


    Chester
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2006
    Posts
    25

    Question

    Would number one be

    -9/4

    Chester
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by Chester
    Would number one be

    -9/4

    Chester
    Yes, thank you for pointing that out (I've changed my last post)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    btw. the roots of \sqrt{m^6} are \pm m^3



    \left(3-2\sqrt7\right)\left(3+2\sqrt7\right)

    Do you know how to use foil?

    F:\;3\times3=9

    O:\;3\times2\sqrt7=6\sqrt{7}

    I:\;-2\sqrt7\times3=-6\sqrt{7}

    L:\;-2\sqrt7\times2\sqrt7=-4\sqrt{7^2}=-4\times7=-28

    now you add:

    F+O+I+L

    9\boxed{-6\sqrt{7}+6\sqrt{7}}-28notice the things in the box cancel out therefore,

    9-28=-19 voila!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jul 2006
    Posts
    25

    Thanks again

    I do know how to use FOIL. I have actually figured out a good deal of these sitting here for the last 12 hours.LOL I used the (a+b)(a-b)=(a^2-b^2). = 3^2-28= -19. I am still working on the last problems.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by Chester
    I do know how to use FOIL. I have actually figured out a good deal of these sitting here for the last 12 hours.LOL I used the (a+b)(a-b)=(a^2-b^2). = 3^2-28= -19. I am still working on the last problems.
    x^{-2}=4

    simplify: \frac{1}{x^2}=4

    reciprocate: x^2=\frac{1}{4}

    find the square root of both sides: x=\sqrt{\frac{1}{4}}

    extend: x=\frac{\sqrt1}{\sqrt4}

    solve: \boxed{x=\frac{1}{2}}

    name the other problems you need help on
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    #2 I'm having trouble understanding what the problem is with #2, since \frac{\sqrt3}{\sqrt{6-1}}=\frac{\sqrt3}{\sqrt{6-1}} than therefore \frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6-1}}=0
    but I assume you've done a typo somewhere...

    #3
    3+\sqrt{x^2-8x}=0

    multiply both sides by 3-\sqrt{x^2-8x} and you'll get: \left(3+\sqrt{x^2-8x}\right)\left(3-\sqrt{x^2-8x}\right)=0\left(3-\sqrt{x^2-8x}\right)

    now think back and you'll remember that (a+b)(a-b)=(a^2-b^2).

    simplify: 3^2-\left(\sqrt{x^2-8x}\right)^2=0

    solve: 9-(x^2-8x)=0

    get rid of parentheses: 9-x^2+8x=0

    rearrange: -x^2+8x+9=0

    now use the quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    substitute: x=\frac{-8\pm\sqrt{8^2-4(-1)(9)}}{2(-1)}

    simplify: x=\frac{-8\pm\sqrt{64+36}}{-2}

    simplify x=\frac{-8\pm\sqrt{100}}{-2}

    use the principle square root: x=\frac{-8+10}{-2}

    subtract: x=\frac{2}{-2}

    divide: \boxed{x=-1}
    Last edited by Quick; August 1st 2006 at 06:41 PM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    alright you have the expression \frac{\sqrt6\times\sqrt{14}}{\sqrt7\times\sqrt3}

    for convenience, let's seperate the fractions: \frac{\sqrt6}{\sqrt3}\times\frac{\sqrt{14}}{\sqrt7  }

    now simplify and you'll get: \sqrt{\frac{6}{3}}\times\sqrt{\frac{14}{7}}

    divide and solve: \sqrt2\times\sqrt2=2
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Jul 2006
    Posts
    25

    Unbelievable

    This is not a hard one but my book is so bad that I could not even get that out of it. Just multiply across.

    Thanks,
    Chester
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Jul 2006
    Posts
    25

    Oops

    This one should be

    <br />
\frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6+1}}<br />

    Would it be difficult to explain this. Do I just divide and then subtract?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by Chester
    This is not a hard one but my book is so bad that I could not even get that out of it. Just multiply across.

    Thanks,
    Chester
    are you asking how to change an equation like this \frac{a}{b}=\frac{x}{y} into this: ax=by

    Well, I'll tell you even if this isn't what you're asking...

    we have equation \frac{a}{b}=\frac{x}{y} so to change this equation the first step would be by multiply both sides by b look below:

    multiply both sides by b: \frac{a}{b}\times b=\frac{x}{y}\times b

    therefore: \frac{ab}{b}=\frac{xb}{y}

    therefore: \frac{a\!\not b}{\not b}=\frac{xb}{y}

    so then we get: a=\frac{xb}{y}

    so multiply both sides by y and you get: ay=xb

    let's test it out:

    \frac{1}{2}=\frac{2}{4}

    change it around and you get: 1\times4=2\times2\quad\rightarrow\quad 4=4

    you see? it works!
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Jul 2006
    Posts
    25

    Thanks for the explanation

    I am a little confused (not that thatís hard to do at this point.) But the problem is

    <br />
\frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6+1}}<br />

    Not

    <br />
\frac{\sqrt3}{\sqrt{6-1}}=\frac{\sqrt3}{\sqrt{6+1}}

    Do I solve this the same way?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

/mathhelpforum @mathhelpforum