Everything looks right. 11.51 by 6.51 are your new dimensions.

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- Jan 24th 2013, 10:17 AM #1

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## Need help with the end of this problem

In order to save space, both the length and width of an advertisement in a newspaper had ot be decreased by the same amount in order to cut its area in half. The original advertisement had a length of 10 centimeters and a width of 15 centimeters. What are the new dimensions of the advertisement?

My work:

(15-x)910-x)=75

150-15x-10x+x^2=75

x^2-25x+75=0

Then set it up in a quadratic and got 21.51 and 3.49. Would that be my answers or would I need to subtract those by 15-x and 10-x to find my new dimensions? If so would it be 15-3.49 and 10-3.49?

Thanks!

- Jan 24th 2013, 11:14 AM #2

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- Jan 24th 2013, 11:22 AM #3

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## Re: Need help with the end of this problem

Hello, skweres1!

In order to save space, both the length and width of an advertisement in a newspaper

had to be decreased by the same amount in order to cut its area in half.

The original advertisement had a length of 10 cm and a width of 15 cms.

What are the new dimensions of the advertisement?

My work:

. . . Good!

Then set it up in a quadratic and got and . . Right!

Would that be my answers? . No, x is the width of the strips trimmed off.

. . or would I need to subtract those by and to find my new dimensions? . Yes!

If so, would they be and ? . Correct!

I assume you made a sketch.

Code:: - - 10 - - : - *---*-------- * - : |///://///////| x : * - + - - - - * - : |///: | : : |///: | : 15 |///: | : : |///: |15-x : |///: | : : |///: | : : |///: | : - *---*---------* - : x : 10-x :