1. ## Reciprocal Roots

Hey guys, having trouble with this problem :-/
Show that the equation $2x^2-5x + 2 = 0$ has roots that are reciprocals of each other. Under what conditions will a quadratic equation in the form $ax^2 + bx + c = 0$ have roots that are reciprocals of each other?

2. Originally Posted by Bajan
Hey guys, having trouble with this problem :-/
you're missing a sign.

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

for the second part, you can expand the general form of: $a(x - a) \left(x - \frac 1a \right) = 0$, and see how $a, b,\mbox{ and }c$ 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 $r_1$ and $r_2$ are the roots of the equation $ax^2 + bx + c = 0$, then $r_1 + r_2 = - \frac ba$ and $r1 \cdot r_2 = \frac ca$.

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

just play around with it. tell us your findings

3. 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

4. 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

5. Hello,Bajan!

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

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

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

. . which can be shown to be reciprocals.

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# if roots of quadratic are reciprocal

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