Answers:
a) 3 in
b) 8 in
c) 3 in
Step-by-step explanation:
Thea area [tex]A[/tex] of a rectangle is defined by the following formula:
[tex]A=(w)(l)[/tex]
Where [tex]w[/tex] is the width and [tex]l[/tex] is the length.
In this situation we are given the area of each rectangle and the equation related, and we need to find the length (then, having the length and the area, we can find the width).
So, in order to find [tex]l[/tex] we have to make each quadratic function equal to zero and find the positive result:
a) [tex]289=36x^{2}-12x+1[/tex]
Making the function equal to zero:
[tex]36x^{2}-12x-288=0[/tex]
Finding the values for [tex]x[/tex] with the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
Where [tex]a=36[/tex], [tex]b=-12[/tex], [tex]c=-288[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]x=\frac{-(-12)\pm\sqrt{(-12)^{2}-4(36)(-288)}}{2(36)}[/tex]
[tex]x=3[/tex] This is the length of rectangle a
b) [tex]1225=25x^{2}-50x+25[/tex]
Making the function equal to zero:
[tex]25x^{2}-50x-1200=0[/tex]
Finding the values for [tex]x[/tex] with the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
Where [tex]a=25[/tex], [tex]b=-50[/tex], [tex]c=-1200[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]x=\frac{-(-50)\pm\sqrt{(-50)^{2}-4(25)(-1200)}}{2(25)}[/tex]
[tex]x=8[/tex] This is the length of rectangle b
c) [tex]289=49x^{2}-56x+16[/tex]
Making the function equal to zero:
[tex]49x^{2}-56x-273=0[/tex]
Finding the values for [tex]x[/tex] with the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
Where [tex]a=49[/tex], [tex]b=-56[/tex], [tex]c=-273[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]x=\frac{-(-56)\pm\sqrt{(-56)^{2}-4(49)(-273)}}{2(49)}[/tex]
[tex]x=3[/tex] This is the length of rectangle c
