One more for you. Last one

[lceman23]

The length of a rectangle is 1mm more than twice the width. If the area is 300 square mm, find the dimensions of the rectangle.
Alright I set this one up as 300= (2w+1)w
0=2w**2 +w-300
(2w-12)(2w+25)
{-25, 12}
2


The answer in the back of my book is 12 and 25. How am I getting this solution set that is wrong. Anyone see where I messed up.

 

[burakaltr]

The length of a rectangle is 1mm more than twice the width. If the area is 300 square mm, find the dimensions of the rectangle.
Alright I set this one up as 300= (2w+1)w
0=2w**2 +w-300
(2w-12)(2w+25)
{-25, 12}
2


The answer in the back of my book is 12 and 25. How am I getting this solution set that is wrong. Anyone see where I messed up.

L=1+2*w

300=w+2*w*w

2*w**2+w-300=

Now find the numbers a*c+b*d= 1

and a*b=2 ;

c*d=-300

Finally

(aw+c)(bw+d)=0

w=-c/a

OR

w=-d/b

 

[lceman23]

I forgot that the 12 needed to be substituted for the 2w+1 and the w. I figured it out from your description thanks a lot.

 

[burakaltr]

I forgot that the 12 needed to be substituted for the 2w+1 and the w. I figured it out from your description thanks a lot.

If you use
roots1,2 = [-b +- sqrt(b**2 - 4 a c ) ] / (2a)

you arrive at r1,2=

-1+-[sqrt(1+2400)] / 4

[-1+-(49)]/4

48/4 = 12

the other soln -50/4 is trivial for it is neg.