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!

Re: Need help with the end of this problem

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

Re: Need help with the end of this problem

Hello, skweres1!

Quote:

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:

. . $\displaystyle (15-x)(10-x)\:=\: 75 \quad\Rightarrow\quad x^2-25x+75\:=\:0$ . Good!

Then set it up in a quadratic and got $\displaystyle 21.51$ and $\displaystyle 3.49$. . Right!

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

. . or would I need to subtract those by $\displaystyle 15-x$ and $\displaystyle 10-x$ to find my new dimensions? . Yes!

If so, would they be $\displaystyle 15-3.49$ and $\displaystyle 10-3.49$? . Correct!

I assume you made a sketch.

Code:

` : - - 10 - - :`

- *---*-------- * -

: |///://///////| x

: * - + - - - - * -

: |///: | :

: |///: | :

15 |///: | :

: |///: |15-x

: |///: | :

: |///: | :

: |///: | :

- *---*---------* -

: x : 10-x :