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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

 Aug 10, 2018
edited by Guest  Aug 10, 2018

Best Answer 

 #3
avatar+30085 
+1

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

 

 

x2+k = 1-kx

 

x2 + kx + k-1 = 0.  (1)

 

The discriminant  of this quadratic is k2-4*1*(k-1) or k2-4k+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.

 Aug 10, 2018
edited by Alan  Aug 11, 2018
 #1
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0

y = x^2 + k

 Aug 10, 2018
 #2
avatar+30085 
+2

The graph above might help (even though it is a little busy!).

 Aug 10, 2018
 #3
avatar+30085 
+1
Best Answer

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

 

 

x2+k = 1-kx

 

x2 + kx + k-1 = 0.  (1)

 

The discriminant  of this quadratic is k2-4*1*(k-1) or k2-4k+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
edited by Alan  Aug 11, 2018

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