Solve algebra problem w^2 + 15w = 324?

Question

Im trying to help my son solve this problem in algebra: A rectangle has a length that is 15ft longer than its width. Its area is 324 sq ft. What is its width?

We got the equations

L = W + 15 and

L * W = 324

that gives

( W + 15 ) W = 324 or....

W^2 + 15W = 324 where "W^2" is "W squared"

That's as far as we can get. Can you help solve for W? PLEASE INCLUDE AN EXPLANATION, as I am trying to help a 14-year old understand this stuff.

Answer 1

Area = length x width

324 = L x W

L = W + 15

324 = (W + 15) x W

W² + 15W - 324 = 0

Hopefully the teacher has covered the quadratic equation or at least factoring quadratics, or it wouldn't be fair to give this problem. Otherwise, you could solve the above equation by trial and error, but that would be tedious. The closed form solution to an equation of the form ax² + bx + c = 0 is: r1,r2 = [-b ± √(b² - 4ac)]/2a

Plug the values in:

W = (-15 ± √(15² - 4(1)(-324))/((2)(1))

W = 12

L = 27

Or if you factor, W² + 15W - 324 = 0 = (W+27)(W-12) and you get W=12

Answer 2

LW=324

W(W+15)=324

W2+15w-324=0

Now you have a quadratic equation.

(w+27)(w-12)=0

so w=-27 or w=12

W=12

since you can't have a negative width.

L=w+15

L=12+15=27

Answer 3

W^2 + 15W = 324

Using the quadratic formula:

W^2 + 15W - 324=0

W=(-15+/-√(15^2-4(-324)))/2

W=(-15+/-39)/2

W=-27 not acceptable

W=12

L = W + 15

L=27

Answer 4

A = 324 ft.²

L = w + 15

A = wL

A = w(w + 15)

324 = w²  + 15w

w² + 15w - 324 = 0

This is a quadratic equation. Using the Quadratic Formula:

[-b ± √(b² - 4ac)] / 2a, where a - 1, b = 15, c = -324, we get

w = 12, w = -27

Since a length can't be negative, w = 12 ft., so

L = 12 + 15

L = 27 ft.

To check:

A = wL

324 = (12)(27)

324 = 324

¯¯¯¯¯¯¯¯¯¯¯

Answer 5

w^2 + 15w = 324

w^2 + 15w - 324 = 0

(w + 27)(w - 12) = 0

w = 12

L = 12 + 15 = 27