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 = 4sqrt{3}, k = -4sqrt{3}
The negative k value is -4sqrt{3}.
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.
