# Math Help - Two Problems I need help with please!

1. ## Two Problems I need help with please!

The first:

When the sum of 396 and four times a positive number is subtracted from the square of the number, the result is 84. Find the number.

What I did:

sqrt(x)-(396+4x)=84

sqrt(x)-396-4x=84

sqrt(x)-4x-312=0

-4x^2-x-97344=0

From here I am stuck, when I put it into the quadratic formula it gives me a negative and we aren't allowed to use imaginary numbers. I'm lost.

Here is the next problem:
A man built a walk of uniform width around a rectangular pool. If the area of the walk is square feet and the dimensions of the pool are feet by feet, how wide is the walk?

a=lw
l=w+4

207=w(w+4)
207=w^2+4w
0=w^2+4w-207

another negative in the quadratic equation...

Any help would be greatly appreciated!

2. ## Re: Two Problems I need help with please!

1.) You used the square root instead of the square.

2.) Consider making four pieces of the walk, which you know the lengths of and set their area equal to 207 ft^2.

3. ## Re: Two Problems I need help with please!

For the first one I squared the equation to get rid of the sqrt(x), did I make an error there?

For the second one, I do not know what you mean here, I am sorry. I drew it out and it still didn't make much sense for me, it's a rectangle within a rectangle

4. ## Re: Two Problems I need help with please!

1- The equation defined by what you wrote is $x^2-(396+4x)=84$ which when expanded equals $x^2-4x-480=0$ and as mentioned you did square root which is the opposite of squaring.Then from there you should be able to solve for $x$ using the quadratic formula which is
$x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$

The answers I got where $x=20$ and $x= -20$ and since x must be positive then the solution is $x=20$ and we ignore the other one.

You went wrong from here:
$sqrt(x)-(396+4x)=84$ <------------- it's $x^2$ not sqrt

$sqrt(x)-396-4x=84$<--------------- it's $x^2$ not sqrt

$sqrt(x)-4x-312=0$<----------------- it's $x^2$ not sqrt and you added 84 instead of subtracting 84 which is not correct (must subtract 84 from both sides to make the equation = 0 so we can solve with quadratic formula)

$-4x^2-x-97344=0$ <--------------- I have no idea what you did here.

I would recommend trying this problem again. Hope this helped!

2- So what i would do is combine the area of the pool + side walk then solve for x.

• Area of pool $4x5=45ft^2$
• Area of side walk 207
• The equation is $(2x+5)(2x+9)=252$ the 2x comes from the unknown width that when added to the dimensions of the pool $5x9$ gives us the equation of the total area of the pool and side walk combined. Here is a photo that might help, because a picture is worth a 1000 words.
• Then expanding our equation will give us a quadratic $4x^2+28x+45=252$ then subtract 252 from both sides to solve for x with the quadratic formula $4x^2+28x-207=0$
• The answers I got where $x= 4.5$ and $x= -8.25$ and since we are dealing with real life measurement problem our answer can not be negative feet. So $x=4.5$ is the only answer. Therefore the width of the side walk is 4.5 ft.

I hope this helped you!

5. ## Re: Two Problems I need help with please!

I now see what you guys were saying with the first problem, and you did a great job of showing me how to set up the second one, thanks!

6. ## Re: Two Problems I need help with please!

no problem, I'm glad to help

7. ## Re: Two Problems I need help with please!

Did you make sketch?

A man built a walk of uniform width around a rectangular pool.
If the area of the walk is 207 square feet, and the dimensions of the pool are 9 feet by 5 feet,
how wide is the walk?

Let $x$ = width of the walk.
Code:
      - *-------------------*
x |   :           :   |
- | - *-----------* - |
: |   |           |   |
5 |   | 5         |   |
: |   |     9     |   |
- | - *-----------* - |
x |   :           :   |
- *-------------------*
: x : - - 9 - - : x :
The total area is: $(2x+9)(2x+5)$
The area of the pool is: $9\cdot 5 \,=\,45$

The area of the walk is 207:
. . $(2x+9)(2x+5) - 45 \:=\:207$

Got it?