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December 19, 2014

December 19, 2014

Posted by **mz deedee** on Friday, November 11, 2011 at 11:00am.

x^4-7x^3+12x^2<0

- precalculus -
**Steve**, Friday, November 11, 2011 at 11:55amy = x^2 * (x^2 - 7x + 12)

= x^2 (x-3)(x-4)

x^2 is always positive, so (x-3)(x-4) must be negative. That is true only for

3 < x < 4 or x in (3,4)

- precalculus -
**Reiny**, Friday, November 11, 2011 at 12:00pmx^2(x^2 - 7x + 12) < 0

x^2(x-3)(x-4) < 0

critical values are x = 0, 3, 4

test for x< 0 , say x = -1

all we care about is the sign, so

(+)(-)(-) < 0 false

test for x between 0 and 3, says x = 1

(+)(-)(-) < 0 false

test for x between 3 and 4, say x = 3.5

(+)(+)(-) < 0 ---> TRUE

test for x > 4 , say x = 5

(+)(+)(+) < 0 false

so the solution is

3 < x < 4

the graph shown here, confirms this

http://www.wolframalpha.com/input/?i=x%5E4-7x%5E3%2B12x%5E2

http://www.wolframalpha.com/input/?i=x%5E4-7x%5E3%2B12x%5E2

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