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Algebra Question

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

Feb 13, 2018

#1
+18328
+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:

Feb 13, 2018
#2
+7598
+1

y   =   2x2 + kx + 6

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

-x + 4  =  2x2 + kx + 6

Add  x  to both sides of the equation.

4   =   2x2 + kx + x + 6

4   =   2x2 + (k + 1)x + 6

Subtract  4  from both sides of the equation.

0   =   2x2 + (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

Feb 13, 2018