Question

One of the roots of kx2 + (1 ¡ 3k)x + (k ¡ 6) = 0 is the negative reciprocal of the other
root. Find k and the two roots.

Answer 1

I know about + and - ... not so sure about ¡
since you know how to type a +, I'll assume you meant
kx2 + (1-3k)x + (k-6) = 0
those roots are
x = ((3k-1) ± √((1-3k)^2 - 4k(k-6)))/(2k) = ((3k-1) ± √(5k^2+18k+1))/(2k)
so now we know that
((3k-1) + √(5k^2+18k+1))/(2k) = -(2k)/((3k-1) - √(5k^2+18k+1))
solve that and you have k=3
now plug that in to get 3x^2-8x-3 = 0
x = 3 or -1/3