# Reciprocal Roots

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• Feb 24th 2008, 10:25 AM
Bajan
Reciprocal Roots
Hey guys, having trouble with this problem :-/
Quote:

Show that the equation $\displaystyle 2x^2-5x + 2 = 0$ has roots that are reciprocals of each other. Under what conditions will a quadratic equation in the form $\displaystyle ax^2 + bx + c = 0$ have roots that are reciprocals of each other?
• Feb 24th 2008, 10:36 AM
Jhevon
Quote:

Originally Posted by Bajan
Hey guys, having trouble with this problem :-/

you're missing a sign.

do you mean $\displaystyle 2x^2 + 5x + 2 = 0$ or $\displaystyle 2x^2 - 5x + 2 = 0$?

for the second part, you can expand the general form of: $\displaystyle a(x - a) \left(x - \frac 1a \right) = 0$, and see how $\displaystyle a, b,\mbox{ and }c$ relate to each other. that's one way. you could also use the quadratic formula to find the roots, and then equate one to the reciprocal of the other and try to solve. things like that.

or perhaps you can use the nature of roots to solve it. if $\displaystyle r_1$ and $\displaystyle r_2$ are the roots of the equation $\displaystyle ax^2 + bx + c = 0$, then $\displaystyle r_1 + r_2 = - \frac ba$ and $\displaystyle r1 \cdot r_2 = \frac ca$.

then plug in $\displaystyle r_2 = \frac 1{r_1}$ and see what happens.

just play around with it. tell us your findings
• Feb 24th 2008, 10:49 AM
CaptainBlack
Quote:

Originally Posted by Bajan
Hey guys, having trouble with this problem :-/

Well for the quadratic 2x^2-5x+2=0 apply the quadratic formula to find the roots.

RonL
• Feb 24th 2008, 10:50 AM
CaptainBlack
Quote:

Originally Posted by Jhevon
you're missing a sign.

The sign was not missing, just bad LaTeX, fixed now.

But it did not matter the same result occurs with either sign.

RonL
• Feb 24th 2008, 01:24 PM
Soroban
Hello,Bajan!

Quote:

Under what conditions will a quadratic equation in the form $\displaystyle ax^2 + bx + c \:=\:0$
have roots that are reciprocals of each other?

An equation with "symmetric" coefficients will have reciprocal roots . . .
. . that is: .$\displaystyle ax^2 + bx + a \:=\:0$

The two roots are: .$\displaystyle x \;=\;\frac{-b + \sqrt{b^2-4a^2}}{2a}\,\text{ and }\,\frac{-b-\sqrt{b^2-4a^2}}{2a}$

. . which can be shown to be reciprocals.