Answer:2 in.
Explanation:
Given
Dimension of photo frame is
![20\ in.* 24\ in.](https://img.qammunity.org/2021/formulas/mathematics/middle-school/w6tzloftomvq6ftt0xpxu9tmd6j3qzyymd.png)
If the photo cover an area of
![320\ in.^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/4ahdfa44527tl9vm6iutpabduh5z2c6kdq.png)
Suppose x be the width of border
Therefore dimension of frame without border is
![A'=(20-2x)(24-2x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/9gbmfkb4c4ba6pl2741gyu7sx0tvb6i3kl.png)
And
must be equal to
![320\ in.^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/4ahdfa44527tl9vm6iutpabduh5z2c6kdq.png)
So,
![\Rightarrow (20-2x)(24-2x)=320](https://img.qammunity.org/2021/formulas/mathematics/middle-school/kzw6dp9jjo6n7ydo3npizni8plcav9buqz.png)
![\Rightarrow (10-x)(12-x)=80](https://img.qammunity.org/2021/formulas/mathematics/middle-school/9cw5lq7mv071s1drn92icnkvgntbv43w2o.png)
![\Rightarrow 120-10x-12x+x^2=80](https://img.qammunity.org/2021/formulas/mathematics/middle-school/s14ek1yvehmxt2qzvtngmggn4ra06tfaus.png)
![\Rightarrow x^2-22x+40=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/q6mxhzaohq2fagvpymnx82dedscx0xfs3u.png)
![\Rightarrow x^2-20x-2x+40=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/l2n5bbwlmv9uy78fvitmfdwztk6sj01o4m.png)
![\Rightarrow (x-2)(x-20)=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vu1kdfwk0vm7uer5i63amvl7bmp0z11nuo.png)
Thus there are two values of x out of which
is not valid because it is not feasible
thus width of border is
![x=2\ in.](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ca42rfnsb893gcw50ma37oatx6f2hzrylz.png)