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**HallsofIvy** Skeeters method, find the roots to the original equation, take the reciprocal, and create the new equation from that is the obvious and simplest method.

Here is a little more "sophisticated" method. Look at the general product (x- a)(x- b)= x^2- (a+b)x+ ab. If, instead, we have (x- 1/a)(x- 1/b) we get x^2- (1/a+ 1/b)x+ 1/ab. You can see that the last term, 1/ab, is just the reciprocal in the original equation, ab. Also 1/a+ 1/b= b/ab+ a/ab= (a+b)/ab so the coefficient of x in the last equation is the coeffient of x in the original equation **divided by** the last term.

Starting from 3x^2+2x-1=0, which is the same as x^2+ (2/3)x- 1/3= 0 (I divided both sides by 3), we can see that the equation having the reciprocals as roots is x^2+ (2/3)/(-1/3)x- 1(1/3)= x^2- 2x- 3= 0.

We can do the same kind of thing for "2a" and "2b". (x- 2a)(x- 2b)= x^2-(2a+2b)+ (2a)(2b)= x^2- 2(a+b)+ 4(ab)= 0. To create an equation that has 2 times the roots as its roots, multiply the coefficient of x by 2 and the final term by 4.

Whatever the roots of 4x^2-5x-2= 0, which is the same as x^2- (5/4)x- 1/2= 0, are, the roots of the equation x^2- (5/2)(2)x- (1/2)(4)= x^2- 5x- 2= 0 are twice that.