Question

The rectangle shown has a length (x+16)cm and width x cm. The square has a length (3x-2)cm. The area of the rectangle is 16cm^2 more than the area of the square. The area of the square is more than 1cm^2. Find the value of x.

Answer 1

Answer:

Step-by-step explanation:

We have the sides for both the rectangle and the square.  The problem says that the area of the rectangle is 16 more than the area of the square.  The area of the rectangle is

(x + 16)(x)

The area of the square is

(3x - 2)(3x - 2)

"The area of the rectangle"      "is"     "16 more than the area of the square"

          x(x + 16)                             =          (3x - 2)(3x - 2) + 16

FOILing the left side and then setting it equal to the FOILing of the right side:

$x^2+16x=9x^2-12x+4+16$ and

$x^2+16x=9x^2-12x+20$

Now we will get everything on the same side of the equals sign, set the polynomial equal to 0, and factor to solve for x:

$8x^2-28x+20=0$

Factor that however you learned best to factor quadratics (the quadratic formula works for a second degree polynomial every time!) to get that

x = 1 or x = 2.5