2x^2-2kx+4=0
2(sqrt3 -1)^2 + 2k(sqrt3-1)+4=0
2(2+ 3-2sqrt3)+2k(sqrt3-1)+4=0
k= (-14+4sqrt3)/(sqrt3 - 1) check that.
if (SQRT(3)-1) is a root of the equation 2x^2 - 2kx+4=0 then k equals what.
2 answers
2(√3 -1)^2 + 2k(√3-1)+4=0
divide by 2
(√3-1)^2 + (√3-1)k + 2 = 0
3-2√3+1 + (√3-1)k +2 = 0
(√3-1)k = 2√3-6
k = (2√3-6)/(√3-1) = -2√3
On the other hand, if you want rational coefficients, then if you think you might have a typo, and the equation is really
2x^2 - 2kx - 4 = 0
then you need the conjugate for a root, so divide by 2 and you have
x^2-kx-2 = 0
(x-(-1+√3))(x-(-1-√3)) = 0
((x-1)-√3)((x-1)+√3) = 0
(x-1)^2 - 3 = 0
x^2-2x+1-3 = 0
x^2-2x-2 = 0
2x^2-4x-4 = 0
and k=2
divide by 2
(√3-1)^2 + (√3-1)k + 2 = 0
3-2√3+1 + (√3-1)k +2 = 0
(√3-1)k = 2√3-6
k = (2√3-6)/(√3-1) = -2√3
On the other hand, if you want rational coefficients, then if you think you might have a typo, and the equation is really
2x^2 - 2kx - 4 = 0
then you need the conjugate for a root, so divide by 2 and you have
x^2-kx-2 = 0
(x-(-1+√3))(x-(-1-√3)) = 0
((x-1)-√3)((x-1)+√3) = 0
(x-1)^2 - 3 = 0
x^2-2x+1-3 = 0
x^2-2x-2 = 0
2x^2-4x-4 = 0
and k=2
1 answer