Word problems...?

Question

I have two word problems I need help on...

1. The square of a certain negative number is equal to five more then one-half of that number. Find the number.
The number is___

2. The width and the length of a rectangle are consecutive even intergers. If the width is decreased by 3 inches, then the area of the resulting rectangle is 24 square inches. Find the
dimensions of the original rectangle.

The dimensions are __inches wide by __ inches long.

Answer 1

Answer 1:
Let the negative number be x

x^2 = (x/2) + 5
x^2 = (x + 10)/2
2x^2 - x - 10 = 0
2x^2 - 5x + 4x - 10 = 0
x(2x - 5) + 2(2x - 5) = 0
(x + 2)(2x - 5) = 0

x + 2 = 0 (or) 2x - 5 = 0
x = -2, (or) x = 5/2

x cannot be positive.

So, x = -2.
The number is -2.

Answer 2:
Let the length be x + 2 inches and let the width be x inches.

(x + 2)(x - 3) = 24
x^2 - x - 6 = 24
x^2 - x - 30 = 0
x^2 - 6x + 5x - 30 = 0
x(x - 6) + 5(x - 6) = 0
(x + 5)(x - 6) = 0

x = 6, -5

x cannot be -5 as the side of a polygon cannot be negative.

So, x = 6
x + 2 = 8

The dimensions are 6 inches wide by 8 inches long.

Answer 2

1.Let the number be x
According to the condition of the problem,
x^2=x/2 +5
=>2x^2=x+10 [multiplying both sides by 2]
=>2x^2-x-10=0
=>2x^2+4x-5x-10=0
=>2x(x+2)-5(x+2)=0
=>(x+2)(2x-5)=0
Rejecting the positive value of x,we getx= -2
Therefore the number is -2

2.Let the width and length of the rectangle be x and x+2 inches
According to the problem,
(x-3)(x+2)=24
=>x^2-x-6=24
=>x^2-x-30=0
=>(x-6)(x+5)=0
rejecting the negative value of x,we get x=6
Therefore the width of the rectangle is 6 inches and the length is 6+2=8 inches
The dimensions are 6 inches wide by 8 inches long

Answer 3

Let n = a certain negative number

n^2 = 5 + (n/2)

2(n^2) = 2(5 + (n/2))

2n^2 = 10 + n

2n^2 - n -10 = 0

(2n - 5) * (n + 2) = 0

=>

2n - 5 = 0
or
n + 2 = 0

2n = 5
n = 5/2

or

n + 2 = 0
n = -2

Although either 5/2 or -2 will satisfy the general conditions of the problem. But, since n is defined as a negative number, the answer is -2.