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