SOLUTION: For what negative value of k is there exactly one solution to this system of equations? y=2x^2+kx+6 y=-x+4

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Question 1085951: For what negative value of k is there exactly one solution to this system of equations?
y=2x^2+kx+6
y=-x+4
Found 2 solutions by rapture, ikleyn:
Answer by rapture(86)   (Show Source): You can put this solution on YOUR website!
Use the discriminant b^2 - 4ac.
a = 2, b = k, c = 6
Set discriminant to 0 and solve for k.
k^2 - 4(2)(6) = 0
k^2 - 8(6) = 0
k^2 - 48 = 0
k^2 = 48
sqrt{k^2} = sqrt{48}
k = sqrt{16}•sqrt{3}
k = 4•sqrt{3}, k = -4•sqrt{3}
The negative k value is -4•sqrt{3}.
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
For what negative value of k is there exactly one solution to this system of equations?
y=2x^2+kx+6
y=-x+4
~~~~~~~~~~~~~~~ 

1.  Reduce the system to one single equation 

    -x+4 = 2x^2 + kx + 6    (1)   (use substitution !).


2.   The system has exactly one solution if and only if the equation (1) has the unique solution.


3.   Simplify the equation (1):

     2x^2 + (k+1) + 2 = 0.


     Its discriminant is  = .


     The equation (1) has the unique solution if and only if the discriminant is zero: 

      = 0  <====>   = 16  <====>  k + 1 = +/-4.


4.   There are 2 solutions for k:  a)  k = 4 - 1 = 3,   and   b)  k = -4 -1 = -5.


Answer. Negative value of k under the question is -5.

Solved.

Ignore other tutor's solution, since it is WRONG.



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