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.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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