solving for an inequality: (x - 9)^2 (x + 4) > 0

Sollie

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\(\displaystyle 6.\, (x\, -\, 9)^2\, (x\, +\, 4)\, >\, 0\)

\(\displaystyle (x\, +\, 3)\, (x\, -\, 3)\, (x\, +\, 4)\, =\, 0\)

\(\displaystyle x\, =\, -3\)

\(\displaystyle x\, =\, 3\)

\(\displaystyle x\, =\, -4\)

While I do realize that this way of finding x intercepts is incorrect and that the two x values are 9 and -4 I still would like to know why is this way incorrect?
 

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While I do realize that this way of finding x intercepts is incorrect and that the two x values are 9 and -4 I still would like to know why is this way incorrect?

\(\displaystyle (x - 9)^2 \ = \ (x - 9)(x - 9) \)

\(\displaystyle (x^2 - 9) \ = \ (x^2 - 3^2) \ = \ (x + 3)(x - 3) \ \ \ or \ \ \ (x - 3)(x + 3)\)
 
While I do realize that this way of finding x intercepts is incorrect and that the two x values are 9 and -4 I still would like to know why is this way incorrect?
Who told you that?
In this case one of your factors is always non-negative. So both factors must be positive.
Therefore, \(\displaystyle \bf{(x-9)^2>0}\) for all \(\displaystyle x\ne 9\) AND \(\displaystyle (x+4)>0\) if \(\displaystyle x>-4\).
Put those together and write out the solution for us.
 
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\(\displaystyle (x - 9)^2 \ = \ (x - 9)(x - 9) \)

\(\displaystyle (x^2 - 9) \ = \ (x^2 - 3^2) \ = \ (x + 3)(x - 3) \ \ \ or \ \ \ (x - 3)(x + 3)\)

Where did \(\displaystyle (x^2 - 9)\) come from?
 
Where did \(\displaystyle (x^2 - 9)\) come from?
The original poster had factored "(x - 9)2" as (x - 3)(x + 3), which is actually (x2 - 92); (x - 9)2 is not at all the same as (x2 - 92), as squaring does not "distribute" over addition.

I suspect that the helper was highlighting that difference. ;)
 
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