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

Guest Feb 13, 2018

#1**+2 **

2x^2+kx+6 = -x+4

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

(2x+ )(x+ ) = 0 the only two factors which multiply to '2' are 1 and 2 or -1 and -2

(2x+1) (x+2) results in TWO values of x (-1/2 and -1)that would satisfy the system of equations...no good.

(2x+2)(x+1) = 0 would result in ONE values for 'x' : (-1) ....we want only ONE (Per the question)

so this would make k+1 = 4 or k=3

Multiply this out to get 2 x^2 + 4x +2 =0 so k+1 = 4 .... or k=3

(2x-2)(x-1) =0 works for only x =1

multiply it out 2x^2 -4x+2 means k+1 = -4 or k=-5

(2x-1)(x-2) has two solutions x = 1/2 and x= 2 no good

SO k = 3 or -5 results in only one solution for the system of equations.

Graphically:

ElectricPavlov Feb 13, 2018

#2**+1 **

y = 2x^{2} + kx + 6

We are given that y = -x + 4 , so we can substitute -x + 4 in for y .

-x + 4 = 2x^{2} + kx + 6

Add x to both sides of the equation.

4 = 2x^{2} + kx + x + 6

4 = 2x^{2} + (k + 1)x + 6

Subtract 4 from both sides of the equation.

0 = 2x^{2} + (k + 1)x + 2

This equation will have one solution when...

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

(k + 1)^{2} - 16 = 0

(k + 1)^{2} = 16

k + 1 = ± 4

k = ± 4 - 1

k = 4 - 1 = 3 or k = -4 - 1 = -5

hectictar Feb 13, 2018