Question 1085951
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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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<pre>
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 {{{(k+1)^2 - 4*2*2}}} = {{{(k+1)^2-16}}}.


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

     {{{(k+1)^2-16}}} = 0  <====>  {{{(k+1)^2}}} = 16  <====>  k + 1 = +/-4.


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


<U>Answer</U>. Negative value of k under the question is -5.
</pre>

Solved.


Ignore other tutor's solution, since it is <U>WRONG</U>.