Question

In the figure, side AB is given by the expression 5x+5x+3, and side BC is 3x+92x−4. The simplified expression for the area of rectangle ABCD is ________ and the restriction on x is_____

Answer 1

Area = side AB * side BC  =  (5x + 5x + 3)(3x + 92x - 4)

= (10x + 3)(95x - 4) =  950x^2 - 40x + 285x - 12

=  950x^2 + 245x - 12  Answer

This cannot be negative  so restriction on x  is  950x^2 + 245x > 12
That is x > 0.0421

Answer 2

Answer:

$A=950x^2+245-12\\\\\{x\in R: (\frac{4}{95} <x<-\frac{3}{10} )\}$

Step-by-step explanation:

The area of a rectangle is given by:

$A=w*h$

$Where:\\\\w=width\\h=height$

Let:

$h=\overline{\rm AB}=5x+5x+3=10x+3\\\\and\\\\w=\overline{\rm BC}=3x+92x-4=95x-4$

So, the area is given by:

$A=(10x+3)*(95x-4)=950x^2-40x+285x-12=950x^2+245x-12$

It wouldn't make sense if the result leads us to an area equal to 0 or to a negative area, therefore:

$A=950x^2+245x-12>0$

Solving for x using the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac} }{2a} =\frac{-245\pm \sqrt{245^2-4(950)(-12)} }{2(950)} =\frac{-245\pm \sqrt{60025+45600} }{1900}\\\\x=\frac{-245\pm \sqrt{105625} }{1900}=\frac{-245\pm 325}{1900}\\\\Hence\\\\x>\frac{4}{95} \\\\and\\\\x<\frac{-3}{10}$

Therefore, the area is given by:

$A=950x^2+245-12\\\\\{x\in R: (\frac{4}{95} <x<-\frac{3}{10} )\}$