Question

the area A enclosed by a recrangle, in square inches, is a function if the length of its sides x and 5 less than x, when measured in inches. this relation is expressed by the formula:A(x)= x^2-5x for x>0find A(6) and solve A(x)=36

Answer 1

we have

A(x)= x^2-5x for x>0



find A(6) and solve A(x)=36

Part 1

Find A(6)

that means

The value of A(x) when the value of x=6

For x=6

substitute

A(6)= 6^2-5(6)

A(6)=36-30

A(6)=6 in^2

Part 2

A(x)=36

Solve for x

substitute

A(x)= x^2-5x

36= x^2-5x

x^2-5x-36=0

Solve using the quadratic formula

we have

a=1

b=-5

c=-36

substitute

`x=\frac{5\pm\sqrt[]{-5^2-4(1)(-36)}}{2}`
`\begin{gathered} x=\frac{5\pm\sqrt[]{169}}{2} \\ \\ x=(5\pm13)/(2) \end{gathered}`

therefore

the solutions for x are

x=9 and x=-4

the solution is x=9 in (because the value of x can not be negative)