Answer:
Part a) The quadratic function is [tex]4x^{2} +20x-39=0[/tex]
Part b) The value of x is [tex]1.5\ in[/tex]
Part c) The photo and frame together are [tex]7\ in[/tex] 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
[tex](6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0[/tex]
Part b) What is the value of x?
Solve the quadratic equation [tex]4x^{2} +20x-39=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem
we have
[tex]4x^{2} +20x-39=0[/tex]
so
[tex]a=4\\b=20\\c=-39[/tex]
substitute in the formula
[tex]x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}[/tex]
[tex]x=\frac{-20(+/-)\sqrt{1,024}} {8}[/tex]
[tex]x=\frac{-20(+/-)32} {8}[/tex]
[tex]x=\frac{-20(+)32} {8}=1.5\ in[/tex] -----> the solution
[tex]x=\frac{-20(-)32} {8}=-6.5\ in[/tex]
Part c) How wide are the photo and frame together?
[tex](4+2x)=4+2(1.5)=7\ in[/tex]
