Question

A square picture with a side length of 4 inches needs to be enlarged. The final area needs to be 81 square inches.

Answer 1

Answer:

$x^2+8x-65=0$

Step-by-step explanation:

Side length of square = 4 inches

Let x be the increase in length

So, New length = x+4

Area of square = $Side^2$

Area of enlarged square = $(x+4)^2$

Using identity : $(a+b)^2=a^2+b^2+2ab$

Area of enlarged square = $x^2+16+8x$

We are given that The final area needs to be 81 square inches.

So,  $x^2+16+8x=81$

$x^2+16+8x-81=0$

$x^2+8x-65=0$

So, Option C is true

Hence  equation can be used to solve for x, the increase in side length of the square in inches is  $x^2+8x-65=0$

Answer 2

Let x be the "enlargement value" of each side :
 Then the enlarged side becomes (x+4) and the square = (x+4)²
The final area should be (x+4)² = 81
Let's expand:
x²+ 8x + 16 = 81
x² + 8x + 16 - 81 = 0
x² + 8x - 65 = 0