Factorise x^2+9x+18 and hence solve x^2+9x+18=0

Given equation is x²+9x+18 = 0

➝ x²+9x = -18

➝ x²+2(9x)/2 = -18

➝ x²+2(x)(9/2) = -18

On adding (9/2)² both sides then

➝ x²+2(x)(9/2)+(9/2)² = -18+(9/2)²

➝ [x+(9/2)]² = -18+(81/4)

➝ [x+(9/2)]² = (-72+81)/4

➝ [x+(9/2)]² = 9/4

➝ x+(9/2) = ±√(9/4)

➝ x+(9/2) = ±3/2

➝ x = (±3/2) -(9/2)

➝ x = (3/2)-(9/2) or -(3/2)-(9/2)

➝ x = (3-9)/2 or (-3-9)/2

➝ x = -6/2 or -12/2

➝ x = -3 or -6

The roots are -3 and -6

or x²+6x+3x+18 = 0

➝ x(x+6)+3(x+6) = 0

➝ (x+6)(x+3) = 0

➝ x+6 = 0 or x+3 = 0

➝ x = -6 or x = -3

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

step-by-step explanation: