or what values of K do the equations y = (1 -K)x + 2, y = - 1/(Kx) have only 1 real solution?
Leypoldon
Answered question
2022-08-02
For what values of K do the equations have only 1 real solution?
Answer & Explanation
optativaspv
Beginner2022-08-03Added 14 answers
For what values of K do the equations y = (1 - K)x +2, y = - have only 1 real solution
. Compare with . We obtain a = K(1-K), b = 2K, c =1. For only one solution the discriminant must be zero. or K = 1/2. If K = 0 then y = -[1/(Kx)] does not make sense. . Your problem may be solved here. However, see if you need thefollowing. If K= 1/2 then y = (x/2)+2 and y = -[2/(x)] are the twofunctions. Equating y = (x/2)+2 = -[2/(x)] or x + 4 = -4/x or or or or x = -2 where y = (x/2) + 2 = (-2/2)+ 2 = 1. Thus they meet at (-2,1). Notice that xy =-2 exists only in quadrants 2 and 4 since whenx is positive y must b negative and when x is negative y must be positive. However, y = (x/2) + 2has a positive slope of 1/2 and y intercept of 2. Hence it does not pass through quadrant 4. Therefore, thosetwo functions must meet in quadrant 2 only. If they meet in only one point then the line must be thetangent to xy = -2. If y = -2/x then . At (-2,1)(dy/dx) = 1/2 same as the slope of the line. Hence, the line is the tangent to the curve.