or what values of K do the equations y = (1 -K)x + 2, y = - 1/(Kx) have only 1 real solution?

Leypoldon

Leypoldon

Answered question

2022-08-02

For what values of K do the equations y = ( 1 K ) x + 2 , y = 1 K x have only 1 real solution?

Answer & Explanation

optativaspv

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
y = ( 1 K ) x + 2 , y = 1 K x
K ( 1 K ) x 2 + 2 K x + 1 = 0. Compare with a x 2 + b x + c = 0. We obtain a = K(1-K), b = 2K, c =1.
For only one solution the discriminant b 2 4 a c must be zero.
4 K 2 4 K ( 1 K ) ( 1 ) = 0 = 8 K 2 4 K = 4 K ( 2 K 1 ) K = 0 or K = 1/2.
If K = 0 then y = -[1/(Kx)] does not make sense. K = 1 / 2.
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 x 2 + 4 x = 4 or x 2 + 4 x + 4 = 0
or ( x + 2 ) 2 = 0 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 ( d y / d x ) = 2 / ( x 2 ). At (-2,1)(dy/dx) = 1/2 same as the slope of the line.
Hence, the line is the tangent to the curve.

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