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# For what negative value of k is there exactly one solution to the system of equations

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For what negative value of k is there exactly one solution to the system of equations

\begin{align*} y &= 2x^2 + kx + 6 \\ y &= -x + 4? \end{align*}

Oct 10, 2019

#1
+19324
+2

First, set the equations =

2x^2+kx+6 = -x+4        collect like terms

2x^2 + x(k+1) +2 = 0    now, if the discriminant = 0 there is only one root

b^2 - 4ac = 0

(k+1)^2 - 4 (2)(2) = 0

k^2 +2k +1 - 16 =0

k^2 + 2k -15 =0

(k+5)(k-3) = 0       so k = -5 or  3      Q asks for the negative value    -5

Oct 10, 2019

#1
+19324
+2

First, set the equations =

2x^2+kx+6 = -x+4        collect like terms

2x^2 + x(k+1) +2 = 0    now, if the discriminant = 0 there is only one root

b^2 - 4ac = 0

(k+1)^2 - 4 (2)(2) = 0

k^2 +2k +1 - 16 =0

k^2 + 2k -15 =0

(k+5)(k-3) = 0       so k = -5 or  3      Q asks for the negative value    -5

ElectricPavlov Oct 10, 2019