Question

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

Answer 1

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

Answer 2

`x^2+9x+18\\\\9x=3x+6x\\18=3\cdot6\\\\x^2+9x+18=x^2+3x+6x+18=x(x+3)+6(x+3)=(x+3)(x+6)\\\\x^2+9x+18=0\iff(x+3)(x+6)=0\iff x+3=0\ or\ x+6=0\\\\x=-3\ or\ x=-6`