Question

Help w/ Quadratic Equations 4?

I know these are two together, please bare with me. Please solve the following. Show your work and be as visual as possible.

Answer 1

The 2 numbers are `=0.13` and `=3.87`

The dimensions of the rectangle is `=16.12m` by `=15.76m` or `=7.88m` by `=32.24m`

Explanation:

Exercise `(4)`

Let the `2` numbers be `x` and ##y

Then,

`x+y=4`

`1/x+1/y=8`

`y=(4-x)`

So,

`1/x+1/(4-x)=8`

`(4-x+x)/(x(4-x))=8`

`1=2x(4-x)=8x-2x^2`

`2x^2-8x+1=0`

`x=(8+-sqrt(64-8))/4=(8+-sqrt56)/4`

`=(8+-2sqrt14)/4`

`=(4+-sqrt14)/2`

`x_1=(4+sqrt14)/2=3.87`, `=>`, `y_1=4-3.87=0.13`

`x_2=(4-sqrt14)/2=0.13`, `=>`, `y_2=4-0.13=3.87`

Exercise `(5)`

Let the dimensions of the rectangle be `xm` and `y m`

The perimeter is `=2x+y=48`

The area is `=xy=254`

Solving for `x` and `y`

`y=254/x`

`2x+254/x=48`

`2x^2+254=48x`

`x^2-24x+127=0`

`x=(24+-sqrt(24^2-4*127))/2`

`=(24+-sqrt68)/2`

`x_1=(24+sqrt68)/2=16.12`

`x_2=(24-sqrt68)/2=7.88`

Therefore,

`x=16.12` , `=>`, `y=48-2*16.12=15.76`

`x=7.88`, `=>`, `y=48-2*7.88=32.24`