Question

Write and solve a quadratic equation for the situation below. Choose the answer that has both an equation that correctly models the situation as well as the correct solution for the situation.
You work for a company that produces custom picture frames. A new customer needs to frame a piece of rectangular artwork with dimensions of 11 x 15 in. They don't want the framed art to be too big so they want to limit its area to 320 in^2. What should the width of the frame be to accommodate their wishes?

A) 4x^2+52x+165=320; x = 2.5

B) 4x^2+52x+165=320; x = 15.5

C) None of the choices are correct.

D) x^2+26x+165=320; x = 5


2) Which quadratic equation below could be used to represent the following situation?

The Cameron's have a triangular baseball pennant hanging from a flagpole in their yard. The pennant has an area of 50 ft^2. The height of the triangle is 20 less than 6 times the length of its base. What are the dimensions of the pennant?

A) 6x^2−20x=50

B) x(6x - 20) + 50 = 0

C) 3x^2−10x=50

D) None of the choices are correct.




3) Which quadratic equation below could be used to represent the following situation?

Your town wants to put a square fountain in the park. The fountain will have a sidewalk around it that is 4 ft wide. The total space that the fountain and sidewalk can use is 800 ft^2. What are the largest dimensions that can be used for the fountain?

A) x^2+16x+64=800

B) x^2+8x+16=800

C) x^2+16=800

D) None of the choices are correct.

Thank you for your help :)

please show and explain your answers and answer all parts. :)

Answer 1

The answer 
the true answer  is A) 4x^2+52x+165=320; x = 2.5 
proof:
since they want to limit its area to 320 in^2
x must be equal 2.5, because 4.(2.5)^2+52(2.5)+165= 25+130+165=320;

the main rule of the area of a triangle is 

A= (base x height )/2
let be b the base and h the height 
so h=20- 6b
so 50= b x( 20-6b) /2, implies 100/b = 20 - 6b
if b=x, we have 100= 20x -6x² this  is equivalent to 6x² - 20x = -100
so the answer is 
D) None of the choices are correct.

the largest dimensions that can be used for the fountain are

B) x^2+8x+16=800

C) x^2+16=800

proof
x^2+8x+16=x^2+8x+4², 4² is the area of the square fountain, and x^2+8x should be the remaining of the area, the total space that the fountain and sidewalk can use is 800, it is  less than 800 ft^2.
Use the same method for. x^2+16=800