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anonymous

  • 5 years ago

2. Suppose x is nonzero, and let x= 1/y a. Show that 3y^2 -16y +1 if x^2-16x+3 and vice versa. b. More generally, show that cy^2 + by + a= 0 if ax^2 + bx + c= 0 and vice versa.

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  1. anonymous
    • 5 years ago
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    this will be a royal pain but i think it is totally doable. first of all i assume the first statement is \[3y^2-16y+1=0\] if \[x^2-16x+3=0\] yes? if so we solve the second equation for x, take its reciprocal, replace it in the first equation and see if we get zero. or just solve them both and see that one solution is the reciprocal of the other.

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    \[x^2-16x=0\] \[x^2-16x=-3\] \[(x-8)^2=-3+64=61\] \[x-8=\sqrt{61}\] or \[x-8=-sqrt{61}\] \[x=8\pm \sqrt{61}\]

  4. anonymous
    • 5 years ago
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    before we solve the other, lets find the reciprocal of this. \[\frac{1}{8+\sqrt{61}}=\frac{1}{8+\sqrt{61}} \times \frac{8-\sqrt{61}}{8-\sqrt{61}}\] \[=\frac{8-\sqrt{61}}{3}\]

  5. anonymous
    • 5 years ago
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    so is this the complete answer!

  6. anonymous
    • 5 years ago
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    so this should be the solution to the other equation. \[3y^2-16y+1=0\] this time may be easier to use formula.

  7. anonymous
    • 5 years ago
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    are we still doing part A

  8. anonymous
    • 5 years ago
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    we have done the following: solved one quadratic. taken the reciprocal of the solution. then we have to see if that is a solution of the other quadratic. which it is because i just checked.

  9. anonymous
    • 5 years ago
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    but we need to write it. we need to write the solution of \[3y^2-16y+1=0\] if you use the quadratic formula you will get \[\frac{8\pm \sqrt{61}}{3}\]

  10. anonymous
    • 5 years ago
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    which is the reciprocal of our first solution. in other words the solution of one equation is the reciprocal of the solution of the other. the second part will be more of a pain because we have to use a, b and c rather than 1, -16 and 3

  11. anonymous
    • 5 years ago
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    oh ok

  12. anonymous
    • 5 years ago
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    actually not really more of a pain because you see that the only difference is that a and c are switched. be is still the same. so it amounts to showing that \[\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] is the reciprocal of \[\frac{-b\pm \sqrt{b^2-4ca}}{2c}\]

  13. anonymous
    • 5 years ago
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    the first being the solution to \[ax^2+bx+c=0\] and the second the solution to \[cx^2+bx+a=0\]

  14. anonymous
    • 5 years ago
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    i can write it if you like. just flip the first one and rationalize the denominator, see that you get the second one.

  15. anonymous
    • 5 years ago
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    this is the answer for part b correct

  16. anonymous
    • 5 years ago
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    i am kind of confused by all the numbers...i dont no what is the answer for part a and part b

  17. anonymous
    • 5 years ago
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    yes. part a is done. solve both equations and see that the solution of one is the reciprocal of the solution to the other.

  18. anonymous
    • 5 years ago
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    ok lets go slow a little. the solutions to \[x^2-16x+3=0\] are \[8 \pm \sqrt{61}\]

  19. anonymous
    • 5 years ago
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    ok

  20. anonymous
    • 5 years ago
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    the reciprocals of these numbers are \[\frac{8\pm \sqrt{61}}{3}\] which you find by taking the reciprocal and rationalizing the denominator.

  21. anonymous
    • 5 years ago
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    and the solutions of \[3y^2-16y+1\] are also these same numbers, namely \[\frac{8\pm \sqrt{61}}{3}\] which you find using the quadratic formula.

  22. anonymous
    • 5 years ago
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    so that completes part a. the solutions of one equation are the reciprocals of the solutions to the other equation.

  23. anonymous
    • 5 years ago
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    now part b asks to show that the solutions of \[cy^2+by+a=0\] are the reciprocals of the solutions of \[ax^2+bx+c=0\] and the only difference in these equations is that the a and c are switched.

  24. anonymous
    • 5 years ago
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    the solutions of \[ax^2+bx+c=0\] are \[\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] which is just the quadratic formula. the solutions to \[cy^2+bx+a\] is exactly the same but with the a and c switched, i.e. \[\frac{-b\pm \sqrt{b^2-4ac}}{2c}\]

  25. anonymous
    • 5 years ago
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    so your only job now is to show that one on of these is the reciprocal of the other. i hope this is making sense at this hour.

  26. anonymous
    • 5 years ago
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    lol i am trying to make it make sense..

  27. anonymous
    • 5 years ago
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    its a lil confusing but I think I will get it

  28. anonymous
    • 5 years ago
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    it may be the logic of it that is confusing you. but your last job is only to show that one of those things is the reciprocal of the other. do it by writing the reciprocal and rationalizing the denominator.

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