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dumb algebra question

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Manager
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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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