a rectangular garden covers 200 yd^2. the length is 8yd longer than the width. find the length and width.

Answer 1

Assuming that your figures are correct (the results are not integers!), area = L x W = 200 sq yds.

But L = W + 8, so 200 = (W + 8)(W) = W² + 8W

In quadratic form, the equation is W² + 8W - 200 = 0.

You'll need to use the quadratic equation to solve (or completing the square). W = -4 ± 6√6.

Answers are L = 18.6969 and W = 10.6969.

Answer 2

Hi Peter,

Let the width= w and length=l

Then, l= 8+w

The area (A) of a rectangle is given by the formula, A=lw

Substituting, we have: 200=(8+w)w=w^2+8w.

This is quadratic. If we move everything to one side, this yields w^2+8w-200=0 which is in the form y=ax^2+bx+c

We need two numbers that multiply to -200 (a*c) and sum to 8 (b). Going through all possibilities, we don't have any. So option two would usually be to apply the quadratic equation:

w= (-(b)±√(b^2-4ac))/2a ----------> w= (-(8)±sqrt((8^2)-4(1)(-200)))/2(1) =

(-8±sqrt(864))/2=(-8±12sqrt(6))/2=-4+6sqrt(6)=w (choose the positive answer)

So the length is 8+w which is l=4+6sqrt(6)

I hope this helps!

Answer 3

200 = w(w + 8)

200 = w² + 8w

0 = w² + 8w - 200

Solve for w using any method you chose.