For what real numbers x is $x^{2} - 6 x + 9$ $x^2-6x+9$ negative?

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For what real numbers x is $x^{2} - 6 x + 9$ $x^2-6x+9$ negative?

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Answer 1 · 2015-04-18T16:45:24

$x^{2} - 6 x + 9 = {(x - 3)}^{2}$ $x^2-6x+9 = (x-3)^2$

This is never negative using real numbers for $x$ $x$

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

Answer 2 · 2015-04-18T16:58:10

The are none.

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

factorize $x^{2} - 6 x + 9$ $x^2-6x+9$

By completing the square,
$x^{2} - 6 x + {(- \frac{6}{2})}^{2} - {(- \frac{6}{2})}^{2} + 9$ $x^2 -6x + (-6/2)^2-(-6/2)^2+9$
= ${(x - 3)}^{2} - 9 + 9$ $(x-3)^2-9+9$
= ${(x - 3)}^{2}$ $(x-3)^2$

Hence,
${(x - 3)}^{2} < 0$ $(x-3)^2<0$

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

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For what real numbers x is $x^{2} - 6 x + 9$ $x^2-6x+9$ negative?

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