How do I solve 2*w^2+12*w+36=174^2?

Question

This is the problem: a parcel of land is 6ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel?

I am stuck here:

2*w^2+12*w+36=174^2

Answer 1

w^2 + (w+6)^2 = 174^4

2w^2 + 12w + 36 = 174^2

2w^2 + 12w - 30240 = 0

w^2 + 6w - 15120 = 0

w^2 - 120w + 126w - 15120 = 0

w(w-120) + 126(w-120) = 0

(w-120)(w+126)=0

w >0 (width cannot be negative)

w = 120

l = w + 6 = 126



The parcel is 120ft by 126ft

Answer 2

Great work! Now, bring all the terms to one side and factor:



2*w^2+12*w+36=174^2 implies



2*w^2+12*w+36 = 30276 implies



2*w^2 + 12*w - 30240 = 0 implies



2 * (w^2 + 6w - 15120) = 0 implies



2 * (w - 120) * (w + 126) = 0 implies



either w = 120 ft or w = -126 ft



Obviously the width cannot be negative. So we end up with:



width = 120 ft

length = 120 ft + 6 ft = 126 ft

Answer 3

a^2 + b^2 = c^2

w^2 + (w+6)^2 = 174^2

w^2 + (w + 6)(w + 6) = 30276

w^2 + (w^2 + 12w + 36) = 30276

2w^2 + 12W + 36 = 30276

2w^2 + 12W - 30240 = 0



use the quadratic formula: x = (-b +- sqrt(b^2 - 4ac))/2a

x = (-12 +- sqrt(144 + 241920))/4

x = (-12 +- sqrt(242064))/4

x = (-12 +- 492)/4

x = (-12 + 492)/4 or x = (-12 - 492)/4

x = 480/4 or x = -504/4

x = 120 or x = -126



the width can't be negative, which means the width is 120 and the height is 120 + 6 = 126.

Answer 4

2*w^2 + 12*W + 36 = 174^2

W^2 +6W + 9 = (174^2)/2 - 18

(w+3)= +/- 123

w = -3 +/- 123

Assuming w is positive, w = 120