On the equation 2kx & # 178; + 4x + 8 = 0 of X, there are two real roots x, Y. question: if two real roots x, y satisfy x-2y = 5, find the value of K. I have listed the ternary equation, ①x﹢y=﹣1／k；②xy=k／4；③x-2y=5 If anyone can help me figure it out,

On the equation 2kx & # 178; + 4x + 8 = 0 of X, there are two real roots x, Y. question: if two real roots x, y satisfy x-2y = 5, find the value of K. I have listed the ternary equation, ①x﹢y=﹣1／k；②xy=k／4；③x-2y=5 If anyone can help me figure it out,

According to Weida's theorem, x + y = - 2 / K, xy = 4 / K, 2 ∵ x-2y = 5, 3 - 3, 3Y = - 2 / k-5y = - 2 / (3K) - 5 / 3x = 5 + 2Y = - 4 / (3K) + 5 / 3xy = 8 / (9K & # 178;) + 10 / (9K) - 25 / 9, the simultaneous formula is 25K & # 178; + 26k-8 = 0, and the root formula k = (- 13 ± 3 √ 41) / 25