4x^2-20x-25=0 is the same problem of course.
You cannot factor this so you can solve with the quadratic formula or
completing the square. Of course you could also use a calculator or computer
algebra system.
Let's complete the square.
4x^2-20x=25
4(x^2-5x)=25
x^2-5x=25/4
(x-5/2)^2=25/4+25/4
(x-5/2)^2=25/2
x-5/2=+ or - 5/square root (2)
x=5/2+5/sqrt(2)
or x=5/2-5/(sqrt2)
We can rationalize the denominator
5/2+5(sqrt(2)/2
5/2-5(sqtrt(2)/2
or
(5+5(sqrt2))/2
or (5-5(sqrt(2))/2
also written as
5(1+sqrt(2))/2
5(1-sqrt(2))/2
r=0 is the solution...
m = 0
Yes, it is the only solution.
Yes.
a = [ 0, 4 ]
root
The two equations are linear.
2*8 - 4*4 = 0, otherwise insert "does not" before equals.
-2
The only solution to this is x=0.
one
When you get an answer like 0=6 it would be no solution.