Q:

The area of a square in square feet isrepresented by 625z^2 − 150z + 9. Find anexpression for the perimeter of thesquare. Then find the perimeter whenz = 15 ft.

Accepted Solution

A:
Answer:Part 1) The expression for the perimeter is [tex]P=4(25z-3)[/tex] or  [tex]P=100z-12[/tex]Part 2) The perimeter when  z = 15 ft. is [tex]P=1,488\ ft[/tex]Step-by-step explanation:Part 1) we have[tex]625z^{2}-150z+9[/tex]Find the roots of the quadratic equationEquate the equation to zero[tex]625z^{2}-150z+9=0[/tex]Complete the squareGroup terms that contain the same variable, and move the constant to the opposite side of the equation[tex]625z^{2}-150z=-9[/tex]Factor the leading coefficient   [tex]625(z^{2}-(150/625)z)=-9[/tex][tex]625(z^{2}-(6/25)z)=-9[/tex]Complete the square. Remember to balance the equation by adding the same constants to each side[tex]625(z^{2}-(6/25)z+(36/2,500))=-9+(36/4)[/tex][tex]625(z^{2}-(6/25)z+(36/2,500))=0[/tex]Rewrite as perfect squares[tex]625(z-6/50)^{2}=0[/tex][tex]z=6/50=0.12[/tex] -----> root with multiplicity 2 soThe area is equal to   [tex]A=625(z-0.12)(z-0.12)=[25(z-0.12)][25(z-0.12)]=(25z-3)^{2}[/tex]The length side of the square is [tex]b=(25z-3)[/tex]thereforeThe perimeter is equal to[tex]P=4b[/tex][tex]P=4(25z-3)[/tex][tex]P=100z-12[/tex]Part 2) Find the perimeter when  z = 15 ft.we have[tex]P=100z-12[/tex]substitute the value of z[tex]P=100(15)-12=1,488\ ft[/tex]