Answer:
Please check the explanation.
Step-by-step explanation:
As we know that the area of a rectangle is defined by multiplying the length by the width.
Given
- rectangular frame length = l = (x+2) units
- rectangular frame width = w = (x-2) units
substituting all the given values in the formula to find the value of x.
![A=l\times w](https://tex.z-dn.net/?f=A%3Dl%5Ctimes%20w)
![96=\left(x+2\right)\times \left(x-2\right)](https://tex.z-dn.net/?f=96%3D%5Cleft%28x%2B2%5Cright%29%5Ctimes%20%5Cleft%28x-2%5Cright%29)
![96=x^2-4](https://tex.z-dn.net/?f=96%3Dx%5E2-4)
![x^2-4=96](https://tex.z-dn.net/?f=x%5E2-4%3D96)
subtract 96 from both sides
![x^2-4-96=96-96](https://tex.z-dn.net/?f=x%5E2-4-96%3D96-96)
![x^2-100=0](https://tex.z-dn.net/?f=x%5E2-100%3D0)
![x^2=100](https://tex.z-dn.net/?f=x%5E2%3D100)
![\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7Dx%5E2%3Df%5Cleft%28a%5Cright%29%5Cmathrm%7B%5C%3Athe%5C%3Asolutions%5C%3Aare%5C%3A%7Dx%3D%5Csqrt%7Bf%5Cleft%28a%5Cright%29%7D%2C%5C%3A%5C%3A-%5Csqrt%7Bf%5Cleft%28a%5Cright%29%7D)
![x=\sqrt{100},\:x=-\sqrt{100}](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B100%7D%2C%5C%3Ax%3D-%5Csqrt%7B100%7D)
![x=10,\:x=-10](https://tex.z-dn.net/?f=x%3D10%2C%5C%3Ax%3D-10)
Putting x = -10 in the length and width will make the length and width negative, which can not be possible.
i.e.
length = l = x+2 = -10+2 = -8 units
width = w = x-2 = -10-2 = -12 units
Therefore, x=-10 must be excluded.
Now, putting the length of x = 10.
i.e.
length = l = 10+2 = 10+2 = 12 units
width = w = x-2 = 10-2 = 8
![A=l\times w](https://tex.z-dn.net/?f=A%3Dl%5Ctimes%20w)
96 = 12 × 8
96 = 96
Therefore, the correct value of x = 10