A small painting has an area of 400 cm^2. The length is 4 more than 2 times the width. Find the dimensions of the painting. Solve by completing the square. Round answers to the nearest tenth of a centimeter.

let x = width

2x + 4 = length
A = lw
A = x(2x + 4)
400 = x(2x + 4)
400 = 2x^2 + 4x
200 = x^2 + 2x
x^2 + 2x = 200
do you know how to complete the square?

no i don't understand still

Yes, I know how to do it from now. Thank you.

Great!

To solve the problem using completing the square, we need to follow these steps:

Step 1: Set up the equation.
Let's assume the width of the painting is "w" cm. According to the problem, the length is 4 more than 2 times the width, which can be expressed as (2w + 4) cm.

The area of the painting can be calculated by multiplying the length and width:
Area = Length × Width
400 cm^2 = (2w + 4) cm × w cm

Step 2: Simplify the equation.
Multiply the terms on the right side:
400 cm^2 = 2w^2 + 4w cm^2

Step 3: Move all the terms to one side to create a quadratic equation.
Rearrange the terms:
2w^2 + 4w - 400 cm^2 = 0

Step 4: Divide the entire equation by the coefficient of w^2 to make the coefficient 1.
Divide each term by 2:
w^2 + 2w - 200 cm^2 = 0

Step 5: Complete the square.
To complete the square, we need to add the square of half the coefficient of w to both sides of the equation. The coefficient of w is 2, so half of that is 1.

Add (1)^2 = 1 to both sides of the equation:

w^2 + 2w + 1 - 200 cm^2 + 1 = 1
(w + 1)^2 - 200 cm^2 + 1 = 0

Step 6: Simplify the equation.
Combine like terms:
(w + 1)^2 - 199 cm^2 = 0

Step 7: Solve for w.
Move the constant term to the other side:
(w + 1)^2 = 199 cm^2

Take the square root of both sides:
w + 1 = ±√199 cm

Subtract 1 from both sides:
w = -1 ± √199 cm

Ignore the negative value since width cannot be negative:
w = √199 - 1 cm

Step 8: Calculate the length.
Using the formula for length we derived earlier, substitute the value of w:
Length = 2w + 4
Length = 2(√199 - 1) + 4

Simplify the expression:
Length = 2√199 + 2

Therefore, the dimensions of the painting are approximately:
Width ≈ √199 - 1 cm
Length ≈ 2√199 + 2 cm