Solving quadratic equation problem

[paulmdrdo1]

please check my answers if they are correct. these problems are even numbered probs in my books that's why I need somebody to check it.

1. solve for x in terms of other symbols

$x^2-2xy-4x-3y^2=0$

using the quadratic formula I get

$x=y+2\pm4\sqrt{y^2+y+1}$

2. what is the width of a strip that must be plowed around a rectangular field 100m long by 60m long wide so that the field will be two-thirds plowed?

let $w=$width around the strip

$60-2w =$ width of the strip
$100-2w =$ length of the strip

so,

$(60-2w)(100-2w)=\frac{2}{3}\times(60)(100)$

the width of the strip is approx. 46.33m

thanks!

 

[MarkFL]

1.) Check the coefficient of the radical.

2.) You have set the problem up incorrectly, and even so obtained the wrong solution to what you set up. You want the inner rectangle, which represents the unplowed portion, to be 1/3 of the total.

 

[paulmdrdo1]

yes, I found my mistake in first problem.

the answer should be $x=y+2\pm2\sqrt{y^2+y+1}$

but in the 2nd how do I set up the problem?

 

[MarkFL]

Set it up very similarly to what you did, only equate the inner rectangle to 1/3 of the total rather than 2/3. Do you see why?

 

[paulmdrdo1]

let $w=$width of the strip

$60-2w =$ width of the inner rectangle
$100-2w =$ length of the inner rectangle

$(60-2w)(100-2w)=$ area of the inner rectangle(unplowed portion)

so,

$(60-2w)(100-2w)=\frac{1}{3}\times(60)(100)$

$w=15.51m$

is my solution and answer now correct?

 

[MarkFL]

Yes, your result is accurate to two decimal places. Unless the instructions were to give an approximate answer, I would give the exact answer:

$$w=10\left(4-\sqrt{6}\right)$$