By solving a quadratic equation, we will see that the border must measure 1 inch.
The dimensions of the paper are 8 inches by 11 inches. If we have a border of width X all around the text, the dimensions of the text will be:
8 in - 2x by 11in - 2x.
The area of this rectangle will be:
A = (8 in - 2x)*(11in - 2x)
And we know that the area must be 54 square inches, then we need to solve:
54 in^2 = (8 in - 2x)*(11in - 2x)
54 in^2 = 88 in^2 - (2x)*8in - (2x)*11in + 4x^2
54 in^2 = 88 in^2 - (38 in)*x + 4x^2
This is a quadratic equation, we can rewrite this as:
0 = 4x^2 - (38 in)*x + 88 in^2 - 54 in^2
0 = 4x^2 - (38 in)*x + 34 in^2
The solutions are given by Bhaskara's formula, these are:
[tex]x = \frac{-(-38 in) \pm \sqrt{(-38 in)^2 - 4*4*(34 in)^2} }{2*4} \\\\x = \frac{-(-38 in) \pm 30 in }{8} \\[/tex]
We only care for the solution that is on the wanted range, so we use the negative sign of the plus-minus sign.
x = (38 in - 30 in)/ = 1 in
So the border must measure 1 inch.
If you want to learn more about quadratic equations, you can read:
https://brainly.com/question/1214333
