For what values of m will the equation x²-2mx+7m-12=0 has reciprocal roots?
Let there be reciprocal roots x1 = a and x2 = 1/a, where these two roots are reciprocal to each other.
Taking these roots as the solutions to a quadratic equation alows us to recreate the quadratic as,
(x - a)(x - 1/a) = 0
Multiplying out gives us,
x^2 - (a + 1/a)x + 1 = 0
Comparing this quadratic with the original x²-2mx+7m-12 = 0, and comparing the coefficient of x and the constant term,
-(a + 1/a) = -2m
1 = 7m - 12 ---> m = 13/7
The equation has reciprocal roots for m = 13/7
(a + 1/a) = 26/7
7a^2 + 7 = 26a
7a^2 - 26a + 7 = 0
Solution here gives the reciprocal roots as a1 = 13/7 + (2/7)sqrt(30) and a2 = 13/7 - (2/7)sqrt(30)