1. ## Problem Solving

I have 6 problems here that I cannot solve... Help me please. These are problems that require factoring... I need the solutions also... Here are the problems..

1.) Find two integers which differ by 5, while their squares differ by 85.

4.) A rectangular swimming pool with dimensions of 11m. by 8m. is built in
rectangular backyard 460m2. If the strip of yard surrounding the is of
uniform width, how wide is the strip?

5.) If the sides of a square are lengthened by 5cm. the area becomes 64cm2.
Find the length of the side of the original square.

6.) The pages of a book are 15cm. by 20cm. Margin of equal width surround the
print on each page and comprise one half the area of the page. Find the
width of the margins.

7.) The sides of a cube is 2cm. more than the side of another cube. The volumes of
the cubes differ by 152cm2. Find the length of the sides of each cube.

8.) The product of two consecutive positive integers is 210. Find the integers.

2. Originally Posted by reiward
8.) The product of two consecutive positive integers is 210. Find the integers.
[tex] x(x+1)=210[tex]
[tex] x^2 + x - 210=0 [tex]

3. x^2+x-210 = 0
that's...
(x-14) | (x+15)
x-14=0 | x+15=0
x=14 | x=-15

They're not both positives?

4. Originally Posted by reiward
5.) If the sides of a square are lengthened by 5cm. the area becomes 64cm2.
Find the length of the side of the original square.
if the original side is x

then (x+5)^2 = 64
or x^2 + 10x + 25=64
or x^2 + 10x -39 =0

dan

5. x^2+10x-39=0
(x-3) (x+13)
x=3 | x=-13

Okay help please.. I always end up having positive and negative answers....

6. Originally Posted by reiward
x^2+x-210 = 0
that's...
(x-14) | (x+15)
x-14=0 | x+15=0
x=14 | x=-15

They're not both positives?

ether way your values are 14 and 15 and 14*15=210
dan

7. Originally Posted by reiward
x^2+10x-39=0
(x-3) (x+13)
x=3 | x=-13

Okay help please.. I always end up having positive and negative answers....
And?
You reject the negative. Thus there is only one answer.

8. Originally Posted by reiward
x^2+10x-39=0
(x-3) (x+13)
x=3 | x=-13

Okay help please.. I always end up having positive and negative answers....
because you are using the quadradic formula you have the "+/- square root" so you will usually have two values...usally only one fits the problem though.
dan

9. alright, i get it.. next problms please

10. Originally Posted by reiward

1.) Find two integers which differ by 5, while their squares differ by 85.
You have,
x-y=5
x^2-y^2=85

Factor the second,
x-y=5
(x-y)(x+y)=85

Substitute,
x-y=5
5(x+y)=85

Thus,
x-y=5
x+y=17

2x=22

Thus,
x=11

Thus,
y=6
Originally Posted by reiward
5.) If the sides of a square are lengthened by 5cm. the area becomes 64cm2.
Find the length of the side of the original square.
The sides of the new square must be 8 cuz 8^2=64.
Thus, the old square must have had 3 cuz 3+5=8.

11. Thank you so much.. Only 3 more left..