anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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.
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[x^2-16x=0\] \[x^2-16x=-3\] \[(x-8)^2=-3+64=61\] \[x-8=\sqrt{61}\] or \[x-8=-sqrt{61}\] \[x=8\pm \sqrt{61}\]

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anonymous
  • anonymous
before we solve the other, lets find the reciprocal of this. \[\frac{1}{8+\sqrt{61}}=\frac{1}{8+\sqrt{61}} \times \frac{8-\sqrt{61}}{8-\sqrt{61}}\] \[=\frac{8-\sqrt{61}}{3}\]
anonymous
  • anonymous
so is this the complete answer!
anonymous
  • anonymous
so this should be the solution to the other equation. \[3y^2-16y+1=0\] this time may be easier to use formula.
anonymous
  • anonymous
are we still doing part A
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \[\frac{8\pm \sqrt{61}}{3}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
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 \[\frac{-b\pm \sqrt{b^2-4ca}}{2c}\]
anonymous
  • anonymous
the first being the solution to \[ax^2+bx+c=0\] and the second the solution to \[cx^2+bx+a=0\]
anonymous
  • anonymous
i can write it if you like. just flip the first one and rationalize the denominator, see that you get the second one.
anonymous
  • anonymous
this is the answer for part b correct
anonymous
  • anonymous
i am kind of confused by all the numbers...i dont no what is the answer for part a and part b
anonymous
  • anonymous
yes. part a is done. solve both equations and see that the solution of one is the reciprocal of the solution to the other.
anonymous
  • anonymous
ok lets go slow a little. the solutions to \[x^2-16x+3=0\] are \[8 \pm \sqrt{61}\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
the reciprocals of these numbers are \[\frac{8\pm \sqrt{61}}{3}\] which you find by taking the reciprocal and rationalizing the denominator.
anonymous
  • anonymous
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.
anonymous
  • anonymous
so that completes part a. the solutions of one equation are the reciprocals of the solutions to the other equation.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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. \[\frac{-b\pm \sqrt{b^2-4ac}}{2c}\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
lol i am trying to make it make sense..
anonymous
  • anonymous
its a lil confusing but I think I will get it
anonymous
  • anonymous
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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