The area of a rectangle is 12 square inches. The length is 5 more than twice the width. Find the length and width

Since the equation for area of a rectangle is A = l*w, and the length is 5 more than twice the width, making it 2w + 5, you have to plug in the quantities.

A = l*w

12 = (2w+5) * w

12 = 2w²+5w

0 = 2w²+5w-12

0 = (2w-3) (w+4)

Since the width can't be a negative number, the answer is:

2w-3 = 0

2w = 3

w = 3/2

And now, since length is 2w+5, you plug in the value of the width into the equation.

2 (3/2) + 5

3+5

8

Let the width of the triangle be x.

Therefore length = 2x + 5. (5 more than twice its width).

Area = L * W = (2x + 5) x = 12.

2x^2 + 5x = 12

2x^2 + 5x - 12 = 0.

This is a quadratic equation. (2) * (-12) = - 24.

We think of two numbers whose product is - 24 and it sum is + 5.

Those two numbers are 8 and - 3. So we replace the middle term of

quadratic with (8 - 3).

2x^2 + 5x - 12 = 0.

2x^2 + 8x-3x - 12 = 0. Factorize.

2x (x + 4) - 3 (x + 4) = 0

(2x-3) (x+4) = 0.

(2x-3) = 0 or (x+4) = 0

2x = 3. x = 0 - 4

x = 3/2 = 1.5 x = - 4. (x can't be negative, since we are solving for lengths)

x = 1.5 is only valid solution.

width = x = 1.5

Length = (2x + 5) = 2*1.5 + 5 = 3 + 5 = 8.

Therefore length = 8, and width = 1.5

Cheers.