Originally Posted by
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.