Prove that x^2 - 6x+13>=4 for real x numbers. Factor the expression A = x^2+4y^2+9-4xy+6x-12y. If x=2y, prove that A=9.
When solving a inequality, we have to determine the interval of numbers for the expression to be positive or negative.
In this case, we'll have to subtract 4 both sides:
x^2 - 6x+13 - 4>=0
x^2 - 6x+ 9 >=0
We remark that we can write the expression above as a binomial raised to square:
(x-3)^2 >= 0
The square of the expression x-3 is always positive for any value of x, excepting x = 3 for the expression is cancelling.
To factor the expresison A, we'll re-write it:
A = x^2+4y^2+9-4xy+6x-12y
A =x^2 - 4xy + 4y^2 + 6x - 12y + 9
We'll group the first 3 factors:
A = (x - 2y)^2 + 6(x - 2y) + 9
We'll put x - 2y = t
A = t^2 - 6t + 9
A = (t - 3)^2
We'll put t = x - 2y
A = (x - 2y - 3)^2
We'll verify if x = 2y the value for A is 9. For this reason, we'll substitute x by 2y:
A = (2y - 2y - 3)^2
We'll eliminate like terms:
A = 9 q.e.d.