*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**

Then,

(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)