Determine the value of k such that g(x) = 3x + k intersects the quadratic function f(x) = 2x^2 - 5x + 3

Question

anonymous · Answer

The answer is supposed to be $$k \ge -5$$

anonymous · Answer

and after I equated I got 2x^2 - 8x + 3-k

.sam. · Answer

b^2 -4ac=0

anonymous · Answer

and then I use the discriminat: (-8)^2 - 4(2)(3-k) = 64 - 8 - 12 but I don't know how to get the k.. is it +4k?

.sam. · Answer

a=2 b=-8 c=3-k

hoblos · Answer

f(x) = g(x)
2x^2 - 5x + 3 = 3x + k
2x^2 -8x +3-k = 0
the two graphs intersect when the quadratic has real solutions

64 - 8(3-k) >= 0
40 +8k >=0
   k >= -5

.sam. · Answer

(-8)^(2)-4(2)(3-k)=0

((64))-4(2)(3-k)=0

(64)-4(2)(3-k)=0

64-4(2)(3-k)=0

64-4(6-2k)=0

64-24+8k=0

40+8k=0

8k=-40

k=-5

anonymous · Answer

ohhh so I multiply the ones with the brackets first?

.sam. · Answer

yes