Answer:
The width of the frame is equal to [tex]8\ cm[/tex]
Step-by-step explanation:
Let
x-------> the width of the frame picture
we know that
[tex](28-2x)(32-2x)=192[/tex]
[tex]896-56x-64x+4x^{2} =192[/tex]
[tex]4x^{2}-120x+704=0[/tex]
Simplify
[tex]x^{2}-30x+176=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]x^{2}-30x+176=0[/tex]
so
[tex]a=1\\b=-30\\c=176[/tex]
substitute in the formula
[tex]x=\frac{30(+/-)\sqrt{(-30)^{2}-4(1)(176)}} {2(1)}[/tex]
[tex]x=\frac{30(+/-)\sqrt{196}} {2}[/tex]
[tex]x=\frac{30(+/-)14} {2}[/tex]
[tex]x=\frac{30+14} {2}=22\ cm[/tex] ----> this solution is not logic
[tex]x=\frac{30-14} {2}=8\ cm[/tex]
The width of the frame is equal to [tex]8\ cm[/tex]
