Question

Show that 3x^2 -2x + 1 is always greater than 0. ( This is an Additional Math Question )

Answer 1

Answer:

3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3

(x-1/3)^2 is greater or equal to zero

3(x-1/3)^2 is greater or equal to zero

and 2/3 is greater than zero

So there sum is greater than zero

Proved

Step-by-step explanation:

3x^2 -2x + 1 =3(x^2-2/3x+1/3)

Consider x^2-2/3x+1/3

Remember that (a-b)^2 =a^2-2ab+b^2

x^2=a^2

a=x

-2/3x= -2*x*b

b=1/3

S0 (x-1/3)^2= x^2-2/3x+1/9

x^2-2/3x+1/3= x^2-2/3x+1/9+1/3-1/9= (x-1/3)^2+2/9

3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3

(x-1/3)^2 is greater or equal to zero

3(x-1/3)^2 is greater or equal to zero

and 2/3 is greater than zero

So there sum is greater than zero

Proved