## 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.

1. 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.

2. anonymous

yes

3. anonymous

$x^2-16x=0$ $x^2-16x=-3$ $(x-8)^2=-3+64=61$ $x-8=\sqrt{61}$ or $x-8=-sqrt{61}$ $x=8\pm \sqrt{61}$

4. anonymous

before we solve the other, lets find the reciprocal of this. $\frac{1}{8+\sqrt{61}}=\frac{1}{8+\sqrt{61}} \times \frac{8-\sqrt{61}}{8-\sqrt{61}}$ $=\frac{8-\sqrt{61}}{3}$

5. anonymous

so is this the complete answer!

6. anonymous

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

are we still doing part A

8. 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.

9. 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 $\frac{8\pm \sqrt{61}}{3}$

10. 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

11. anonymous

oh ok

12. 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 $\frac{-b\pm \sqrt{b^2-4ca}}{2c}$

13. anonymous

the first being the solution to $ax^2+bx+c=0$ and the second the solution to $cx^2+bx+a=0$

14. anonymous

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

this is the answer for part b correct

16. anonymous

i am kind of confused by all the numbers...i dont no what is the answer for part a and part b

17. 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.

18. anonymous

ok lets go slow a little. the solutions to $x^2-16x+3=0$ are $8 \pm \sqrt{61}$

19. anonymous

ok

20. anonymous

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

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

so that completes part a. the solutions of one equation are the reciprocals of the solutions to the other equation.

23. 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.

24. 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. $\frac{-b\pm \sqrt{b^2-4ac}}{2c}$

25. 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.

26. anonymous

lol i am trying to make it make sense..

27. anonymous

its a lil confusing but I think I will get it

28. 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.