Find the square root of 25/289
1 answer:
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Xy=11
x+y=-26
x+y=-26
minus x both sides
y=-26-x
sub for y
x(-26-x)=11
-26x-x^2=11
add 26x+x^2 both sides
x^2+26x+11=0
ax^2+bx+c=0
x=
a=1
b=26
c=11
x=
x=
x=
x=
x=
x=
or 
aprox
x=-0.420195 or -25.5698
those are the numbers
x+y=-26
x+y=-26
minus x both sides
y=-26-x
sub for y
x(-26-x)=11
-26x-x^2=11
add 26x+x^2 both sides
x^2+26x+11=0
ax^2+bx+c=0
x=
a=1
b=26
c=11
x=
x=
x=
x=
x=
x=
aprox
x=-0.420195 or -25.5698
those are the numbers
For this case we have the following expression:
we observe that we have a polynomial of the second degree.
The constant term of the polynomial, is that coefficient that is not written accompanying a variable.
We have then that the constant term of the quadratic function is:
6
Answer:
the constant of the function is:
6