dumb algebra question : GMAT Problem Solving (PS)
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# dumb algebra question

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Manager
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15 Aug 2011, 20:07
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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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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:
$$(10-x)(9-x) < 9$$
$$90-9x-10x+x^2 < 9$$
$$x^2-19x+81<0$$

$$ax^2+bx+c=0$$
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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16 Aug 2011, 06:31
where did you get 1/2 and sqrt 37?? wht am i missing??
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16 Aug 2011, 06:44
DeeptiM wrote:
where did you get 1/2 and sqrt 37?? wht am i missing??

$$ax^2+bx+c=0$$
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:
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16 Aug 2011, 06:48
I am soo sorry Fluke..I know this formula & knw hw to solve it..

Re: dumb algebra question   [#permalink] 16 Aug 2011, 06:48
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