a-maths!!!!!!!!

g(x)
=-x²-2x-2
=-(x²+2x+1²-1²)-2
=-(x²+2x+1)-2+1
=-(x+1)²-1//

since (x+1)²>0
so -(x+1)²<0

and also -(x+1)²-2<0 => g(x)<0//

(b)試求k可取的0兩個數數值,使得f(x)-kg(x)=0有等根。

f(x)-kg(x)=0,有等根

4x+1-k(-x²-2x-2)=0

4x+1+kx²+2kx+2k=0
kx²+2(k+2)x+(1+2k)=0

Δ=0

[2(k+2)]²-4(k)(1+2k)=0

4(k²+4k+4)-4k-8k²=0
4k²+16k+16-4k-8k²=0
-4k²+12k+16=0
k²-3k-4=0
(k-4)(k+1)=0
k=4 or k=-1 //