\(x^2+15x+k\) with a=1,b=15 and c=k we want double root so \(\sqrt{b^2-4ac}=0 <=> \sqrt{15^2-4k}=0 <=> \sqrt{15^2} =\sqrt{4k}=0 <=> k=225/4\)
\(\frac{ \sqrt{1500}}{4π}=\frac{38.7298334621}{π}=\frac{38.7298334621}{3,14...}\)~3.08202222031
Hi my friend
I found it thank you.
by which theorem?
I don't know why don't show the letters...
1)3x+3y=10
2)(-9x - 9y)/-3 = -30/-3 <=> 3x + 3y = 10 so 1) = 2) so have solution for every x,y
Yes you are right There is no x and y!!!
x = - 1 - y and replace this in the other 3(- 1 - y)=4 - 3y <=> - 3 - 3y = 4 - 3y so -3 = 4 is false and we dont have solutions
13+3x=13+2x <=> 13-13=2x-3x <=> 0=-x <=> x=0/-1 <=> x=0
