15. An irregular pentagon has side lengths of (x + 3), (2x - 4), (4x + 5),
1 answer:
You might be interested in
Factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4
first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x=
so
1x^2-2x+4=0
a=1
b=-2
c=4
x=
x=
x=
we have
and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x=
i=
x=
x=
x=
or x=
(those are the 2 numbers)
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4
first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x=
1x^2-2x+4=0
a=1
b=-2
c=4
x=
x=
x=
we have
therefor there are no real solutions
but if you want to solve fully
x=
i=
x=
x=
x=