Quadratic Equation Reciprocals help?

[The Man]

Well you might have seen this question around:

Show that the equation 2x²-5x+2=0 has roots that are reciprocals of each other. Under what conditions will a quadratic equation in the form ax²+bx+c=0 have roots that are reciprocals of each other?

Anyways, I DO NOT WANT THE ANSWER, I WANT CLARIFICATION. I can get the answer because I'm sure I can just Google it and get the answer but I don't want that because on the tests, I wont have Google and Yahoo Answers to help me.

Anyways, so I get the roots, flip them over and proved they were reciprocals of each other. The second half of the question is what I am stumped on, now before you answer, don't solve it! Could you please instead, push me in the right direction of the answer? Give me some strategies or small hints to it? Thank you very much in advance!

-The Man

 

[soroban]

Hello, The Man!

Show that the equation \(\displaystyle 2x^2-5x+2\:=\:0\) has roots that are reciprocals of each other.

Under what conditions will an equation of the form \(\displaystyle ax^2+bx+c\:=\:0\)
have roots that are reciprocals of each other?

\(\displaystyle \text{The roots of }ax^2+bx+c\:=\:0\text{ are, of course: }\;\frac{-b + \sqrt{b^2-4ac}}{2a}\:\text{ and }\:\frac{-b - \sqrt{b^2-4ac}}{2a}\)


\(\displaystyle \text{If they are reciprocals, then we have: }\;\frac{-b + \sqrt{b^2-4ac}}{2a} \;=\;\frac{2a}{-b - \sqrt{b^2-4ac}}\)

\(\displaystyle \text{Simplify that and determine what the conditions are.}\)

 

[The Man]

Soroban, would this involve radicals?

EDIT: I think to simplify I would need to use radicals, but I just can't grasp on how to do it with more than two variables and no numbers. Correct me if I'm wrong.

EDIT: Solved! I just took two reciprocal roots and worked backwards and found it by myself *pats on back*. Thanks Soroban, your example helped me get there!

 

[Denis]

Did you simplify Soroban's equation to get a = c ?

 

[The Man]

Did you simplify Soroban's equation to get a = c ?

No, I did not but I feel I should now since I have come to another stump! I was going to stop my work their and put it as the answer but I tried to negative fractions being the roots as reciprocals of each other, and as it turns out it is not a or c squared plus one :S. So now I can prove the a=c part (But still do not know how to simplify the quadratic equation) but now I am stuck at how to confirm the B value. Could someone please explain how to simplify Soroban's equation, and head me off in the right direction to finding the B value.

NOTE: I found out that the B value is the numerator squared + the denominator squared, but that was my pure fluke, but if I had only the quad equation, how would I know how to do this since I wouldn't have the root :|

 

[Denis]

\(\displaystyle \text{If they are reciprocals, then we have: }\;\frac{-b + \sqrt{b^2-4ac}}{2a} \;=\;\frac{2a}{-b - \sqrt{b^2-4ac}}\)

Use crisscross multiplication:
4a^2 = b^2 - (b^2 - 4ac) : remember that (a - b)(a + b) = a^2 - b^2
4a^2 = b^2 - b^2 + 4ac
a = c

 

[stapel]

Well you might have seen this question around:

Show that the equation 2x²-5x+2=0 has roots that are reciprocals of each other. Under what conditions will a quadratic equation in the form ax²+bx+c=0 have roots that are reciprocals of each other?

Yes: we saw this question here, three days previous to your posting. :wink:

Eliz.

 

[The Man]

Use crisscross multiplication:
4a^2 = b^2 - (b^2 - 4ac) : remember that (a - b)(a + b) = a^2 - b^2
4a^2 = b^2 - b^2 + 4ac
a = c

4a²=b²-(b²-4ac)
I really have no idea how you got those numbers!

But anyways, after much thinking and trial and error I got the solution! I worked backwards from the roots to the quadratic equation, did various numbers and it all worked out to prove that a=c. The B value can be any number, and I have proved that as well through examples! Thanks for all the help guys, now I think I'm going to do some more math homework Than maybe if I have time answer other peoples questions.