Answer:
I believe the answer is the system has one solution. Both lines have the same y-intercept. And the solution is the intersection of the 2 lines.
We start with 56/154. Dividing each number by two, we have:
28/77
Dividing each number by seven, we have:
4/11
As the greatest common factor of 4 and 11 is 1, this fraction is in simplest form. Thus, 56/154 is equal to 4/11.
Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.
Because we know 3 and -1 (multiplicity 2) are both roots, the last root
is such that we can write

There are a few ways we can go about finding
, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be
.
Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

so the missing root is 3.
Other things we could have tried that spring to mind:
- three rounds of division, dividing the quartic polynomial by
, then by
twice, and noting that the remainder upon each division should be 0
- rational root theorem
notice that, if Q is the midpoint of PR, that simply means that PQ = QR = 43.
![\bf \boxed{P}\underset{\leftarrow 9x-31 \to }{\stackrel{\stackrel{\downarrow }{43}}{\rule[0.35em]{10em}{0.25pt}}Q\stackrel{43}{\rule[0.35em]{10em}{0.25pt}}}\boxed{R} \\\\\\ 9x-31=43+43\implies 9x-31=86\implies 9x=117 \\\\\\ x=\cfrac{117}{9}\implies x=13](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7BP%7D%5Cunderset%7B%5Cleftarrow%209x-31%20%5Cto%20%7D%7B%5Cstackrel%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7B43%7D%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7DQ%5Cstackrel%7B43%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D%7D%5Cboxed%7BR%7D%0A%5C%5C%5C%5C%5C%5C%0A9x-31%3D43%2B43%5Cimplies%209x-31%3D86%5Cimplies%209x%3D117%0A%5C%5C%5C%5C%5C%5C%0Ax%3D%5Ccfrac%7B117%7D%7B9%7D%5Cimplies%20x%3D13)
Answer:
Both answers will give an area of 2400 ft2
but with x=60 we have lawn dimensions -60 ft by -40 ft so this is out
x = 10 ft width for the sidewalk
Check: New lawn dimensions
(80-2x)(60-2x) = 60(40) = 2400 ft^2
Step-by-step explanation:
Draw a diagram:
We have a rectangle inside a rectangle.
The larger outside rectangle is the original lawn: 80ft by 60 ft with area 4800 ft2
The smaller inside rectangle is (80-2x)by(60-2x) where x is width of the new sidewalk.
Area of new lawn is 2400 ft^2
(80-2x)(60-2x) = 2400
4800 - 160x - 120x + 4x2 = 2400
4x2 - 280x + 2400 = 0
Factor out a 4
x2 - 70x + 600 = 0
(x-60)(x-10) = 0
x = 60 ft or x = 10 ft