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Homework Help: Quick help needed on a simple problem (not simple4me)

  1. Mar 20, 2005 #1
    quick help needed on a "simple" problem (not simple4me)

    hey everyone i hope ur all doing great! ok tomorrows my math final and i have a slight dilemma on the review sheet and i would appreciate it a lot if someone helped me with it!


    here it is:

    [tex]z^4+75=28z^2[/tex]

    ok so i first brought the 75 to the other side which gave me:

    [tex]z^4=28z^2-75[/tex]

    and now im stuck! i dont know if i should go back and rewrite z^4 as [tex]z^2z^2[/tex] or what??

    any help is appreciated!! thanks!!
     
  2. jcsd
  3. Mar 20, 2005 #2

    xanthym

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    Hint: Make the substitution {x = z2} and solve the resulting quadratic equation with "x" for the variable (for the first step). Can you see what to do after that??


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    Last edited: Mar 20, 2005
  4. Mar 20, 2005 #3
    let me try and see if i understood what u meant! i'll be back in a bit! thanks btw
     
  5. Mar 20, 2005 #4
    ok so i took ur advice and if i understood it correctly this is what i did with it, however i checked my answer and it didn't check...

    [tex]z^4+75=28z^2[/tex]
    [tex]z^4+75=28x; x=z^2[/tex]
    [tex]z^4=28x-75[/tex]
    [tex]x^2=28x-75, x=z^2[/tex] so [tex]x^2=z^4[/tex]
    [tex]x^2-28x-75=0[/tex]

    then i plugged this into the quadtratic and got this:

    [tex]x=14\pm\sqrt{271}[/tex]

    so i got this approx.:

    [tex]x\approx30.460[/tex]​
    OR
    [tex]x\approx-2.460[/tex]​

    but those answers dont work though... ??
     
  6. Mar 20, 2005 #5
    Should that really be [itex]x^2 - 28x - 75 =0[/itex]? I like positive [itex]75[/itex]s more...
     
  7. Mar 20, 2005 #6
    but does that make a difference when it comes to the quadtratic formula?
     
  8. Mar 20, 2005 #7

    xanthym

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    YES!! The correct equation is given below. Can you see how to factor it??
    (Or you can also use the quadratic formula.)
    x2 - 28*x + 75 = 0


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    Last edited: Mar 20, 2005
  9. Mar 20, 2005 #8
    holy crap ur right!! when i brought the 28x over, i should've ADDED a negative -75 to both sides, which would make it +75 yep ur right!

    brb!! :) :)
     
  10. Mar 20, 2005 #9
    x = 25 OR x = 3


    is that right???
     
  11. Mar 20, 2005 #10

    xanthym

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    CORRECT!!
    Remember, solving the equation in "x" is just the first step. You then must place the "x" solutions back into your original {x = z2} substitution equation and solve for "z" (for both "x" solutions).
    x2 - 28*x + 75 = 0

    ~~
     
    Last edited: Mar 20, 2005
  12. Mar 20, 2005 #11
    yea i went back and did that but it wasn't that obvious for me at first :D

    ok so in the solution of x=3, since z^2=x, that means that z=the square root of x, or in this case, the square root of 3.

    and i plugged that value in the original formula of z^4+75=28z^2 and it worked.

    and i did the same thing for the other value of x, which was 25 and did the same thing!

    so basically, z=/sqrt(3) or z=5 (because the sqrt of 25=5)

    thanks a lot man!

    p.s. both of those are plus or minus btw!!
     
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