What is the vakue of k where kx^2+x+1 and x^2+kx+1 share only one root?
k = -2 is the only answer
-2x^2 + x + 1 = 0 has 2 single roots -1/2 and 1
x^2 -2x + 1 = 0 has one double root at 1
Thus they share only one root, x = 1
k = 1 transforms both equations into the same equation : x^2 + x + 1 = 0
This single equation has 2 different complex roots. (You can never have a double complex root for a quadratic equation). Therefore k = 1 does not satisfy "share only one root"
The equation you have to solve is :
k^2 - 1 = +/-sqrt(1 - 4k) +/-k*sqrt(4k^2 - 1)
where k cannot be 0 as this results in roots x = -1 and x = +/-i from the original equations.
There are 4 cases to consider but you will find that after squaring both sides twice, all 4 cases result in the same equation:
(k + 2)(k - 1)^2 = 0 giving k = -2 or k = 1
k = 1 is an extraneous root, as shown above, leaving only k = -2 which needs to be verified as well since we are solving a radical equation.