Math: Help me with this Word Problem?

Question

Can anyone here solve this problem?

The length of a rectangle is 3 inches greater than its width. Its area is 70 square inches. What are its dimensions?

Answer 1

length = L

width = L-3

Area=L*(L-3)=70

L^2-3L=70

L^2-3L-70=0

solve for L with quadratic formula:

L=-7 and L=10

-7 is not a dimension so the length is 10 and the width is 10-3=7

To check, 10*7=70

Answer 2

say width = x

now length = x+3

area of rectangle = width * length = x(x+3) = x^2 + 3x

now, question says, area = 70, which means

x2+3x=70

or x2+3x-70 = 0

roots will be 10 and -7 for that quadratic equation and above step equals to (x+10)(x-7) = 0

=> x can be either -10 (width can't be negative, so  rule this out)

or x can be +7 (valid width)

so, answer is width = x = 7 inches and length = x+3 = 7+3 = 10 inches

Answer 3

If we call width w we can then call the length w + 3.  You multiply them together to get area so

w * l = 70 or w * (w + 3) = 70 or w^2 + 3w = 70 or w^2 + 3w - 70 = 0,

now use quadratic equation to solve, you must be studying that to have a question like this.

Answer 4

x(x + 3) = 70

x^2 + 3x - 70 = 0

(x - 7)(x + 10) = 0

x = 7, -10

It can't be -10, so the dimensions are 7 x 10

Answer 5

L * W = 70.  But L = W + 3, so

(W + 3) * W = 70

W^2 +3W - 70 = 0

(W + 10)(W - 7) = 0

W = 7 (ignore the negative answer in this context)

L = 10