kx + y = 1

y + -x^2 + k

In the system of equation above, k is a constant. When the equations are graphed in the xy - plane, the graph intersect at exactly two points. Which of the following CANNOT be the value of k?

A. 3

B. 2

C. 1

D. 0

Guest Aug 10, 2018

edited by
Guest
Aug 10, 2018

#3**+1 **

Alternatively, you could do it algebraically. Equate the two expressions for y:

x^{2}+k = 1-kx

x^{2} + kx + k-1 = 0. (1)

The discriminant of this quadratic is k^{2}-4*1*(k-1) or k^{2}-4k+4. This is zero when k = 2.

Since there is only a single solution to equation (1) when the discriminant is zero, k = 2 doesn’t allow the original two equations to intersect in more than one point.

Alan
Aug 10, 2018

#3**+1 **

Best Answer

Alternatively, you could do it algebraically. Equate the two expressions for y:

x^{2}+k = 1-kx

x^{2} + kx + k-1 = 0. (1)

The discriminant of this quadratic is k^{2}-4*1*(k-1) or k^{2}-4k+4. This is zero when k = 2.

Since there is only a single solution to equation (1) when the discriminant is zero, k = 2 doesn’t allow the original two equations to intersect in more than one point.

Alan
Aug 10, 2018