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For what real numbers x is #x^2-6x+9# negative?

Answer 1

Kennedy Adams

#x^2-6x+9 = (x-3)^2#

This is never negative using real numbers for #x#

(It is #0# for #x=3# and positive for every other real number value of #x#.)

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Answer 2

Abigail Allen

There aren't any.

Method To figure that out you let # x^2-6x+9<0#

factorize #x^2-6x+9 #

By completing the square, #x^2 -6x + (-6/2)^2-(-6/2)^2+9# =#(x-3)^2-9+9 # =# (x-3)^2#

Hence, #(x-3)^2<0#

Since the square of any real number is always positive there are really no values of #x# for which that expression is negative!

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Answer 3

Jace Anderson

To find the real numbers for which ( x^2 - 6x + 9 ) is negative, we need to solve the inequality ( x^2 - 6x + 9 < 0 ). This quadratic expression factors as ( (x - 3)^2 ). Since the square of any real number is nonnegative, ( (x - 3)^2 ) is always nonnegative, and hence can never be negative. Therefore, there are no real numbers ( x ) for which ( x^2 - 6x + 9 ) is negative.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.