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Veronika [31]
3 years ago
9

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

of the frame is x inches. The combined area of the photo and frame is 63 square inches.
a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame. You must show your work to receive credit.
b) What is the value of x? You must show your work to receive credit.
c) How wide are the photo and frame together? You must show your work to receive credit.

USATESTPREP WORK!! PLEASE HELP!!
Mathematics
1 answer:
Ede4ka [16]3 years ago
5 0

Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

so

a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

x=\frac{-20(+/-)\sqrt{1,024}} {8}

x=\frac{-20(+/-)32} {8}

x=\frac{-20(+)32} {8}=1.5\ in  -----> the solution

x=\frac{-20(-)32} {8}=-6.5\ in

Part c) How wide are the photo and frame together?

(4+2x)=4+2(1.5)=7\ in

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