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dumb algebra question

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Manager
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Joined: 03 Aug 2011
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Concentration: General Management, Entrepreneurship
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dumb algebra question [#permalink] New post 15 Aug 2011, 20:07
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0% (00:00) correct 100% (00:00) wrong based on 1 sessions
saw this on a practice test as part of a tough problem

how did they solve this inequality?

(10-x)(9-x) < 9

they got

x > 6

thanks
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Re: dumb algebra question [#permalink] New post 16 Aug 2011, 02:16
pinchharmonic wrote:
saw this on a practice test as part of a tough problem

how did they solve this inequality?

(10-x)(9-x) < 9

they got

x > 6

thanks


Please post the question in its entirety without shortening or rewording the question.

For the question above as is:
\((10-x)(9-x) < 9\)
\(90-9x-10x+x^2 < 9\)
\(x^2-19x+81<0\)

For quadratic equation:
\(ax^2+bx+c=0\)
Roots will be:
\(\alpha,\beta=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Likewise, for
\(x^2-19x+81<0\)

Roots:
\((x-\frac{1}{2}(19+\sqrt{37}))(x-\frac{1}{2}(19-\sqrt{37}))<0\)

Or

\(\frac{1}{2}(19-\sqrt{37}) < x < \frac{1}{2}(19+\sqrt{37})\)

Or Approximation for rounded integral values of "x":

\(6 < x < 13\)
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Re: dumb algebra question [#permalink] New post 16 Aug 2011, 06:31
where did you get 1/2 and sqrt 37?? wht am i missing??
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Re: dumb algebra question [#permalink] New post 16 Aug 2011, 06:44
DeeptiM wrote:
where did you get 1/2 and sqrt 37?? wht am i missing??


For quadratic equation:
\(ax^2+bx+c=0\)
Roots will be:
\(\alpha,\beta=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)------------------------1

Use the formula in equation 1 for:
\(x^2-19x+81=0\)

\(a=coefficient \hspace{2} of \hspace{2} x^2=1\)

\(b=coefficient \hspace{2} of \hspace{2} x=-19\)

\(c=constant=81\)

Please solve it to find \(\alpha, \beta\)

Reference:
Look for: "Solving equations of degree 2 : QUADRATIC" under
algebra-101576.html#p787276

External Reference:
http://www.purplemath.com/modules/quadform.htm
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Re: dumb algebra question [#permalink] New post 16 Aug 2011, 06:48
I am soo sorry Fluke..I know this formula & knw hw to solve it..

may b a bad day :(
Re: dumb algebra question   [#permalink] 16 Aug 2011, 06:48
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